Integration by parts

In calculus, and more generally in mathematical analysis, integration by parts or partial integration is a process that finds the integral of a product of functions in terms of the integral of the product of their derivative and antiderivative. It is frequently used to transform the antiderivative of a product of functions into an antiderivative for which a solution can be more easily found. The rule can be thought of as an integral version of the product rule of differentiation; it is indeed derived using the product rule.

The integration by parts formula states:

{\begin{aligned}\int _{a}^{b}u(x)v'(x)\,dx&={\Big [}u(x)v(x){\Big ]}_{a}^{b}-\int _{a}^{b}u'(x)v(x)\,dx\\&=u(b)v(b)-u(a)v(a)-\int _{a}^{b}u'(x)v(x)\,dx.\end{aligned}}

Or, letting $u=u(x)$ and $du=u'(x)\,dx$ while Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle v=v(x)} and $dv=v'(x)\,dx,$ the formula can be written more compactly: Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int u\,dv\ =\ uv-\int v\,du.}

The former expression is written as a definite integral and the latter is written as an indefinite integral. Applying the appropriate limits to the latter expression should yield the former, but the latter is not necessarily equivalent to the former.

Mathematician Brook Taylor discovered integration by parts, first publishing the idea in 1715.^[1]^[2] More general formulations of integration by parts exist for the Riemann–Stieltjes and Lebesgue–Stieltjes integrals. The discrete analogue for sequences is called summation by parts.

Theorem

Product of two functions

The theorem can be derived as follows. For two continuously differentiable functions $u(x)$ and $v(x)$ , the product rule states:

{\Big (}u(x)v(x){\Big )}'=u'(x)v(x)+u(x)v'(x).

Integrating both sides with respect to $x$ ,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \Big(u(x)v(x)\Big)'\,dx = \int u'(x)v(x)\,dx + \int u(x)v'(x) \,dx, }

and noting that an indefinite integral is an antiderivative gives

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u(x)v(x) = \int u'(x)v(x)\,dx + \int u(x)v'(x)\,dx,}

where we neglect writing the constant of integration. This yields the formula for integration by parts:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int u(x)v'(x)\,dx = u(x)v(x) - \int u'(x)v(x) \,dx, }

or in terms of the differentials Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=u'(x)\,dx} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle dv=v'(x)\,dx, \quad}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int u(x)\,dv = u(x)v(x) - \int v(x)\,du.}

This is to be understood as an equality of functions with an unspecified constant added to each side. Taking the difference of each side between two values Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = a} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = b} and applying the fundamental theorem of calculus gives the definite integral version: Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{a}^{b}u(x)v'(x)\,dx=u(b)v(b)-u(a)v(a)-\int _{a}^{b}u'(x)v(x)\,dx.} The original integral Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int uv' \, dx} contains the derivative $v'$ ; to apply the theorem, one must find $v$ , the antiderivative of $v'$ , then evaluate the resulting integral Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int vu' \, dx.}

Validity for less smooth functions

It is not necessary for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v} to be continuously differentiable. Integration by parts works if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} is absolutely continuous and the function designated Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v'} is Lebesgue integrable (but not necessarily continuous).^[3] (If Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle v'} has a point of discontinuity then its antiderivative Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v} may not have a derivative at that point.)

If the interval of integration is not compact, then it is not necessary for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} to be absolutely continuous in the whole interval or for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v'} to be Lebesgue integrable in the interval, as a couple of examples (in which Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v} are continuous and continuously differentiable) will show. For instance, if

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u(x)= e^x/x^2, \, v'(x) =e^{-x}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} is not absolutely continuous on the interval $[1, \infty)$ , but nevertheless:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_1^\infty u(x)v'(x)\,dx = \Big[u(x)v(x)\Big]_1^\infty - \int_1^\infty u'(x)v(x)\,dx}

so long as $\left[u(x)v(x)\right]_{1}^{\infty }$ is taken to mean the limit of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u(L)v(L)-u(1)v(1)} as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L\to\infty} and so long as the two terms on the right-hand side are finite. This is only true if we choose Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v(x)=-e^{-x}.} Similarly, if

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u(x)= e^{-x},\, v'(x) =x^{-1}\sin(x)}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v'} is not Lebesgue integrable on the interval $[1, \infty)$ , but nevertheless

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_1^\infty u(x)v'(x)\,dx = \Big[u(x)v(x)\Big]_1^\infty - \int_1^\infty u'(x)v(x)\,dx} with the same interpretation.

One can also easily come up with similar examples in which Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} and $v$ are not continuously differentiable.

Further, if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)} is a function of bounded variation on the segment Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b],} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi(x)} is differentiable on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b],} then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{a}^{b}f(x)\varphi'(x)\,dx=-\int_{-\infty}^{\infty} \widetilde\varphi(x)\,d(\widetilde\chi_{[a,b]}(x)\widetilde f(x)),}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d(\chi_{[a,b]}(x)\widetilde f(x))} denotes the signed measure corresponding to the function of bounded variation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \chi_{[a,b]}(x)f(x)} , and functions Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \widetilde f, \widetilde \varphi} are extensions of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f, \varphi} to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \R,} which are respectively of bounded variation and differentiable.^{[citation needed]}

Product of many functions

Integrating the product rule for three multiplied functions, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u(x)} , $v(x)$ , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle w(x)} , gives a similar result:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_a^b u v \, dw \ =\ \Big[u v w\Big]^b_a - \int_a^b u w \, dv - \int_a^b v w \, du.}

In general, for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} factors

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(\prod_{i=1}^n u_i(x) \right)' \ =\ \sum_{j=1}^n u_j'(x)\prod_{i\neq j}^n u_i(x), }

which leads to

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[ \prod_{i=1}^n u_i(x) \right]_a^b \ =\ \sum_{j=1}^n \int_a^b u_j'(x) \prod_{i\neq j}^n u_i(x). }

Visualization

File:Integration by parts v2.svg

Graphical interpretation of the theorem. The pictured curve is parametrized by the variable t.

Consider a parametric curve Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (x, y) = (f(t), g(t))} . Assuming that the curve is locally one-to-one and integrable, we can define Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} x(y) &= f(g^{-1}(y)) \\ y(x) &= g(f^{-1}(x)) \end{align}}

The area of the blue region is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_1=\int_{y_1}^{y_2}x(y) \, dy}

Similarly, the area of the red region is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_2=\int_{x_1}^{x_2}y(x)\,dx}

The total area A₁ + A₂ is equal to the area of the bigger rectangle, x₂y₂, minus the area of the smaller one, x₁y₁:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \overbrace{\int_{y_1}^{y_2}x(y) \, dy}^{A_1}+\overbrace{\int_{x_1}^{x_2}y(x) \, dx}^{A_2}\ =\ \biggl.x \cdot y(x)\biggl|_{x_1}^{x_2} \ =\ \biggl.y \cdot x(y)\biggl|_{y_1}^{y_2}} Or, in terms of t,

\int _{t_{1}}^{t_{2}}x(t)\,dy(t)+\int _{t_{1}}^{t_{2}}y(t)\,dx(t)\ =\ {\biggl .}x(t)y(t){\biggl |}_{t_{1}}^{t_{2}}

Or, in terms of indefinite integrals, this can be written as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x\,dy + \int y \,dx \ =\ xy} Rearranging: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x\,dy \ =\ xy - \int y \,dx} Thus integration by parts may be thought of as deriving the area of the blue region from the area of rectangles and that of the red region.

This visualization also explains why integration by parts may help find the integral of an inverse function f⁻¹(x) when the integral of the function f(x) is known. Indeed, the functions x(y) and y(x) are inverses, and the integral ∫ x dy may be calculated as above from knowing the integral ∫ y dx. In particular, this explains use of integration by parts to integrate logarithm and inverse trigonometric functions. In fact, if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is a differentiable one-to-one function on an interval, then integration by parts can be used to derive a formula for the integral of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f^{-1}} in terms of the integral of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} . This is demonstrated in the article, Integral of inverse functions.

Applications

Finding antiderivatives

Integration by parts is a heuristic rather than a purely mechanical process for solving integrals; given a single function to integrate, the typical strategy is to carefully separate this single function into a product of two functions u(x)v(x) such that the residual integral from the integration by parts formula is easier to evaluate than the single function. The following form is useful in illustrating the best strategy to take:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int uv\,dx = u \int v\,dx - \int\left(u' \int v\,dx \right)\,dx.}

On the right-hand side, u is differentiated and v is integrated; consequently it is useful to choose u as a function that simplifies when differentiated, or to choose v as a function that simplifies when integrated. As a simple example, consider:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{\ln(x)}{x^2}\,dx\,.}

Since the derivative of ln(x) is ⁠1/x⁠, one makes (ln(x)) part u; since the antiderivative of ⁠1/x²⁠ is −⁠1/x⁠, one makes ⁠1/x²⁠ part v. The formula now yields:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\frac{\ln(x)}{x^2}\,dx = -\frac{\ln(x)}{x} - \int \biggl(\frac1{x}\biggr) \biggl(-\frac1{x}\biggr)\,dx\,.}

The antiderivative of −⁠1/x²⁠ can be found with the power rule and is ⁠1/x⁠.

Alternatively, one may choose u and v such that the product u′ (∫v dx) simplifies due to cancellation. For example, suppose one wishes to integrate:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\sec^2(x)\cdot\ln\Big(\bigl|\sin(x)\bigr|\Big)\,dx.}

If we choose u(x) = ln(|sin(x)|) and v(x) = sec²x, then u differentiates to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\tan x}} using the chain rule and v integrates to tan x; so the formula gives:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\sec^2(x)\cdot\ln\Big(\bigl|\sin(x)\bigr|\Big)\,dx = \tan(x)\cdot\ln\Big(\bigl|\sin(x)\bigr|\Big)-\int\tan(x)\cdot\frac1{\tan(x)} \, dx\ .}

The integrand simplifies to 1, so the antiderivative is x. Finding a simplifying combination frequently involves experimentation.

In some applications, it may not be necessary to ensure that the integral produced by integration by parts has a simple form; for example, in numerical analysis, it may suffice that it has small magnitude and so contributes only a small error term. Some other special techniques are demonstrated in the examples below.

Polynomials and trigonometric functions

In order to calculate

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I=\int x\cos(x)\,dx\,,}

let: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{3} u &= x\ &\Rightarrow\ &&du &= dx \\ dv &= \cos(x)\,dx\ &\Rightarrow\ && v &= \int\cos(x)\,dx = \sin(x) \end{alignat}}

then:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int x\cos(x)\,dx & = \int u\ dv \\ & = u\cdot v - \int v \, du \\ & = x\sin(x) - \int \sin(x)\,dx \\ & = x\sin(x) + \cos(x) + C, \end{align}}

where C is a constant of integration.

For higher powers of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} in the form

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x^n e^x\,dx,\ \int x^n\sin(x)\,dx,\ \int x^n\cos(x)\,dx\,,}

repeatedly using integration by parts can evaluate integrals such as these; each application of the theorem lowers the power of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} by one.

Exponentials and trigonometric functions

An example commonly used to examine the workings of integration by parts is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I=\int e^x\cos(x)\,dx.}

Here, integration by parts is performed twice. First let

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{3} u &= \cos(x)\ &\Rightarrow\ &&du &= -\sin(x)\,dx \\ dv &= e^x\,dx\ &\Rightarrow\ &&v &= \int e^x\,dx = e^x \end{alignat}}

then:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int e^x\cos(x)\,dx = e^x\cos(x) + \int e^x\sin(x)\,dx.}

Now, to evaluate the remaining integral, we use integration by parts again, with:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{alignat}{3} u &= \sin(x)\ &\Rightarrow\ &&du &= \cos(x)\,dx \\ dv &= e^x\,dx\,&\Rightarrow\ && v &= \int e^x\,dx = e^x. \end{alignat}}

Then:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int e^x\sin(x)\,dx = e^x\sin(x) - \int e^x\cos(x)\,dx.}

Putting these together,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int e^x\cos(x)\,dx = e^x\cos(x) + e^x\sin(x) - \int e^x\cos(x)\,dx.}

The same integral shows up on both sides of this equation. The integral can simply be added to both sides to get

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2\int e^x\cos(x)\,dx = e^x\bigl[\sin(x)+\cos(x)\bigr] + C,}

which rearranges to

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int e^x\cos(x)\,dx = \frac{1}{2}e^x\bigl[\sin(x)+\cos(x)\bigr] + C'}

where again Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C} (and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C' = \frac{C}{2}} ) is a constant of integration.

A similar method is used to find the integral of secant cubed.

Functions multiplied by unity

Two other well-known examples are when integration by parts is applied to a function expressed as a product of 1 and itself. This works if the derivative of the function is known, and the integral of this derivative times Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} is also known.

The first example is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \ln(x) dx} . We write this as:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I=\int\ln(x)\cdot 1\,dx\,.}

Let:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u = \ln(x)\ \Rightarrow\ du = \frac{dx}{x}} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle dv = dx\ \Rightarrow\ v = x}

then:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int \ln(x)\,dx & = x\ln(x) - \int\frac{x}{x}\,dx \\ & = x\ln(x) - \int 1\,dx \\ & = x\ln(x) - x + C \end{align} }

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C} is the constant of integration.

The second example is the inverse tangent function Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \arctan(x)} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I=\int\arctan(x)\,dx.}

Rewrite this as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\arctan(x)\cdot 1\,dx.}

Now let:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u = \arctan(x)\ \Rightarrow\ du = \frac{dx}{1+x^2}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle dv = dx\ \Rightarrow\ v = x}

then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int\arctan(x)\,dx & = x\arctan(x) - \int\frac{x}{1+x^2}\,dx \\[8pt] & = x\arctan(x) - \frac{\ln(1+x^2)}{2} + C \end{align} }

using a combination of the inverse chain rule method and the natural logarithm integral condition.

LIATE rule

The LIATE rule is a rule of thumb for integration by parts. It involves choosing as u the function that comes first in the following list:^[4]

L – logarithmic functions: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ln(x),\ \log_b(x),} etc.
I – inverse trigonometric functions (including hyperbolic analogues): Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \arctan(x),\ \arcsec(x),\ \operatorname{arsinh}(x),} etc.
A – algebraic functions (such as polynomials): Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^2,\ 3x^{50},} etc.
T – trigonometric functions (including hyperbolic analogues): Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin(x),\ \tan(x),\ \operatorname{sech}(x),} etc.
E – exponential functions: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^x,\ 19^x,} etc.

The function which is to be dv is whichever comes last in the list. The reason is that functions lower on the list generally have simpler antiderivatives than the functions above them. The rule is sometimes written as "DETAIL", where D stands for dv and the top of the list is the function chosen to be dv. An alternative to this rule is the ILATE rule, where inverse trigonometric functions come before logarithmic functions.

To demonstrate the LIATE rule, consider the integral

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x \cdot \cos(x) \,dx.}

Following the LIATE rule, u = x, and dv = cos(x) dx, hence du = dx, and v = sin(x), which makes the integral become Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \cdot \sin(x) - \int 1 \sin(x) \,dx,} which equals Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \cdot \sin(x) + \cos(x) + C.}

In general, one tries to choose u and dv such that du is simpler than u and dv is easy to integrate. If instead cos(x) was chosen as u, and x dx as dv, we would have the integral

{\frac {x^{2}}{2}}\cos(x)+\int {\frac {x^{2}}{2}}\sin(x)\,dx,

which, after recursive application of the integration by parts formula, would clearly result in an infinite recursion and lead nowhere.

Although a useful rule of thumb, there are exceptions to the LIATE rule. A common alternative is to consider the rules in the "ILATE" order instead. Also, in some cases, polynomial terms need to be split in non-trivial ways. For example, to integrate

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x^3 e^{x^2} \,dx,}

one would set

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u = x^2, \quad dv = x \cdot e^{x^2} \,dx,}

so that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du = 2x \,dx, \quad v = \frac{e^{x^2}}{2}.}

Then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x^3 e^{x^2} \,dx = \int \left(x^2\right) \left(xe^{x^2}\right) \,dx = \int u \,dv = uv - \int v \,du = \frac{x^2 e^{x^2}}{2} - \int x e^{x^2} \,dx.}

Finally, this results in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x^3 e^{x^2} \,dx = \frac{e^{x^2}\left(x^2 - 1\right)}{2} + C.}

Integration by parts is often used as a tool to prove theorems in mathematical analysis.

Wallis product

The Wallis infinite product for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac{\pi}{2} & = \prod_{n=1}^\infty \frac{ 4n^2 }{ 4n^2 - 1 } = \prod_{n=1}^\infty \left(\frac{2n}{2n-1} \cdot \frac{2n}{2n+1}\right) \\[6pt] & = \Big(\frac{2}{1} \cdot \frac{2}{3}\Big) \cdot \Big(\frac{4}{3} \cdot \frac{4}{5}\Big) \cdot \Big(\frac{6}{5} \cdot \frac{6}{7}\Big) \cdot \Big(\frac{8}{7} \cdot \frac{8}{9}\Big) \cdot \; \cdots \end{align}}

may be derived using integration by parts.

Gamma function identity

The gamma function is an example of a special function, defined as an improper integral for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z > 0 } . Integration by parts illustrates it to be an extension of the factorial function:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \Gamma(z) & = \int_0^\infty e^{-x} x^{z-1} dx \\[6pt] & = - \int_0^\infty x^{z-1} \, d\left(e^{-x}\right) \\[6pt] & = - \Biggl[e^{-x} x^{z-1}\Biggl]_0^\infty + \int_0^\infty e^{-x} d\left(x^{z-1}\right) \\[6pt] & = 0 + \int_0^\infty \left(z-1\right) x^{z-2} e^{-x} dx\\[6pt] & = (z-1)\Gamma(z-1). \end{align} }

Since

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Gamma(1) = \int_0^\infty e^{-x} \, dx = 1,}

when Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z} is a natural number, that is, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z = n \in \mathbb{N} } , applying this formula repeatedly gives the factorial: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Gamma(n+1) = n!}

Use in harmonic analysis

Integration by parts is often used in harmonic analysis, particularly Fourier analysis, to show that quickly oscillating integrals with sufficiently smooth integrands decay quickly. The most common example of this is its use in showing that the decay of function's Fourier transform depends on the smoothness of that function, as described below.

Fourier transform of derivative

If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is a Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} -times continuously differentiable function and all derivatives up to the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} th one decay to zero at infinity, then its Fourier transform satisfies

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\mathcal{F}f^{(k)})(\xi) = (2\pi i\xi)^k \mathcal{F}f(\xi),}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f^{(k)}} is the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} th derivative of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} . (The exact constant on the right depends on the convention of the Fourier transform used.) This is proved by noting that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dy} e^{-2\pi iy\xi} = -2\pi i\xi e^{-2\pi iy\xi},}

so using integration by parts on the Fourier transform of the derivative we get

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} (\mathcal{F}f')(\xi) &= \int_{-\infty}^\infty e^{-2\pi iy\xi} f'(y)\,dy \\ &=\left[e^{-2\pi iy\xi} f(y)\right]_{-\infty}^\infty - \int_{-\infty}^\infty (-2\pi i\xi e^{-2\pi iy\xi}) f(y)\,dy \\[5pt] &=2\pi i\xi \int_{-\infty}^\infty e^{-2\pi iy\xi} f(y)\,dy \\[5pt] &=2\pi i\xi \mathcal{F}f(\xi). \end{align}}

Applying this inductively gives the result for general Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} . A similar method can be used to find the Laplace transform of a derivative of a function.

Decay of Fourier transform

The above result tells us about the decay of the Fourier transform, since it follows that if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f^{(k)}} are integrable then

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vert\mathcal{F}f(\xi)\vert \leq \frac{I(f)}{1+\vert 2\pi\xi\vert^k}, \text{ where } I(f) = \int_{-\infty}^\infty \Bigl(\vert f(y)\vert + \vert f^{(k)}(y)\vert\Bigr) \, dy.}

In other words, if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} satisfies these conditions then its Fourier transform decays at infinity at least as quickly as 1/|ξ|^k. In particular, if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k \geq 2} then the Fourier transform is integrable.

The proof uses the fact, which is immediate from the definition of the Fourier transform, that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vert\mathcal{F}f(\xi)\vert \leq \int_{-\infty}^\infty \vert f(y) \vert \,dy.}

Using the same idea on the equality stated at the start of this subsection gives

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vert(2\pi i\xi)^k \mathcal{F}f(\xi)\vert \leq \int_{-\infty}^\infty \vert f^{(k)}(y) \vert \,dy.}

Summing these two inequalities and then dividing by 1 + |2 $π$ ξ^k| gives the stated inequality.

Use in operator theory

One use of integration by parts in operator theory is that it shows that the −∆ (where ∆ is the Laplace operator) is a positive operator on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L^2} (see L^p space). If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is smooth and compactly supported then, using integration by parts, we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \langle -\Delta f, f \rangle_{L^2} &= -\int_{-\infty}^\infty f''(x)\overline{f(x)}\,dx \\[5pt] &=-\left[f'(x)\overline{f(x)}\right]_{-\infty}^\infty + \int_{-\infty}^\infty f'(x)\overline{f'(x)}\,dx \\[5pt] &=\int_{-\infty}^\infty \vert f'(x)\vert^2\,dx \geq 0. \end{align}}

Other applications

Determining boundary conditions in Sturm–Liouville theory
Deriving the Euler–Lagrange equation in the calculus of variations

Repeated integration by parts

Considering a second derivative of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v} in the integral on the LHS of the formula for partial integration suggests a repeated application to the integral on the RHS: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int u v''\,dx = uv' - \int u'v'\,dx = uv' - \left( u'v - \int u''v\,dx \right).}

Extending this concept of repeated partial integration to derivatives of degree $n$ leads to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int u^{(0)} v^{(n)}\,dx &= u^{(0)} v^{(n-1)} - u^{(1)}v^{(n-2)} + u^{(2)}v^{(n-3)} - \cdots + (-1)^{n-1}u^{(n-1)} v^{(0)} + (-1)^n \int u^{(n)} v^{(0)} \,dx.\\[5pt] &= \sum_{k=0}^{n-1}(-1)^k u^{(k)}v^{(n-1-k)} + (-1)^n \int u^{(n)} v^{(0)} \,dx. \end{align}}

This concept may be useful when the successive integrals of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v^{(n)}} are readily available (e.g., plain exponentials or sine and cosine, as in Laplace or Fourier transforms), and when the $n$ th derivative of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} vanishes (e.g., as a polynomial function with degree Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (n-1)} ). The latter condition stops the repeating of partial integration, because the RHS-integral vanishes.

In the course of the above repetition of partial integrations the integrals Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int u^{(0)} v^{(n)}\,dx \quad} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \quad \int u^{(\ell)} v^{(n-\ell)}\,dx \quad} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \quad \int u^{(m)} v^{(n-m)}\,dx \quad\text{ for } 1 \le m,\ell \le n} get related. This may be interpreted as arbitrarily "shifting" derivatives between Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} within the integrand, and proves useful, too (see Rodrigues' formula).

Tabular integration by parts

The essential process of the above formula can be summarized in a table; the resulting method is called "tabular integration"^[5] and was featured in the film Stand and Deliver (1988).^[6]

For example, consider the integral

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x^3 \cos x \,dx \quad} and take Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \quad u^{(0)} = x^3, \quad v^{(n)} = \cos x.}

Begin to list in column A the function Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u^{(0)} = x^3} and its subsequent derivatives Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u^{(i)}} until zero is reached. Then list in column B the function Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v^{(n)} = \cos x} and its subsequent integrals Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v^{(n-i)}} until the size of column B is the same as that of column A. The result is as follows:

# i	Sign	A: derivatives Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u^{(i)}}	B: integrals Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v^{(n-i)}}
0	+	Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^3}	Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos x}
1	−	Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3x^2}	Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin x}
2	+	Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 6x}	Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\cos x}
3	−	Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 6}	Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\sin x}
4	+	Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0}	Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos x}

The product of the entries in row $i$ of columns A and B together with the respective sign give the relevant integrals in step $i$ in the course of repeated integration by parts. Step $i = 0$ yields the original integral. For the complete result in step $i > 0$ the $i$ th integral must be added to all the previous products ( $0 \leq j < i$ ) of the $j$ th entry of column A and the $(j + 1)$ st entry of column B (i.e., multiply the 1st entry of column A with the 2nd entry of column B, the 2nd entry of column A with the 3rd entry of column B, etc. ...) with the given $j$ th sign. This process comes to a natural halt, when the product, which yields the integral, is zero ( $i = 4$ in the example). The complete result is the following (with the alternating signs in each term):

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \underbrace{(+1)(x^3)(\sin x)}_{j=0} + \underbrace{(-1)(3x^2)(-\cos x)}_{j=1} + \underbrace{(+1)(6x)(-\sin x)}_{j=2} +\underbrace{(-1)(6)(\cos x)}_{j=3}+ \underbrace{\int(+1)(0)(\cos x) \,dx}_{i=4: \;\to \;C}.}

This yields

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \underbrace{\int x^3 \cos x \,dx}_{\text{step 0}} = x^3\sin x + 3x^2\cos x - 6x\sin x - 6\cos x + C. }

The repeated partial integration also turns out useful, when in the course of respectively differentiating and integrating the functions Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u^{(i)}} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v^{(n-i)}} their product results in a multiple of the original integrand. In this case the repetition may also be terminated with this index $i.$ This can happen, expectably, with exponentials and trigonometric functions. As an example consider

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int e^x \cos x \,dx. }

# i	Sign	A: derivatives Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u^{(i)}}	B: integrals Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v^{(n-i)}}
0	+	Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^x}	Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \cos x}
1	−	$e^{x}$	Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin x}
2	+	Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^x}	Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\cos x}

In this case the product of the terms in columns A and B with the appropriate sign for index $i = 2$ yields the negative of the original integrand (compare rows $i = 0$ and $i = 2$ ).

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \underbrace{\int e^x \cos x \,dx}_{\text{step 0}} = \underbrace{(+1)(e^x)(\sin x)}_{j=0} + \underbrace{(-1)(e^x)(-\cos x)}_{j=1} + \underbrace{\int(+1)(e^x)(-\cos x) \,dx}_{i= 2}. }

Observing that the integral on the RHS can have its own constant of integration Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C'} , and bringing the abstract integral to the other side, gives:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2 \int e^x \cos x \,dx = e^x\sin x + e^x\cos x + C', }

and finally:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int e^x \cos x \,dx = \frac 12 \left(e^x ( \sin x + \cos x ) \right) + C,}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C = \frac{C'}{2}} .

Higher dimensions

Integration by parts can be extended to functions of several variables by applying a version of the fundamental theorem of calculus to an appropriate product rule. There are several such pairings possible in multivariate calculus, involving a scalar-valued function u and vector-valued function (vector field) V.^[7]

The product rule for divergence states:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nabla \cdot ( u \mathbf{V} ) \ =\ u\, \nabla \cdot \mathbf V \ +\ \nabla u\cdot \mathbf V.}

Suppose Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Omega} is an open bounded subset of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \R^n} with a piecewise smooth boundary Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Gamma=\partial\Omega} . Integrating over Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Omega} with respect to the standard volume form Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d\Omega} , and applying the divergence theorem, gives:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{\Gamma} u \mathbf{V} \cdot \hat{\mathbf n} \,d\Gamma \ =\ \int_\Omega\nabla\cdot ( u \mathbf{V} )\,d\Omega \ =\ \int_\Omega u\, \nabla \cdot \mathbf V\,d\Omega \ +\ \int_\Omega\nabla u\cdot \mathbf V\,d\Omega,}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{\mathbf n}} is the outward unit normal vector to the boundary, integrated with respect to its standard Riemannian volume form Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d\Gamma} . Rearranging gives:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_\Omega u \,\nabla \cdot \mathbf V\,d\Omega \ =\ \int_\Gamma u \mathbf V \cdot \hat{\mathbf n}\,d\Gamma - \int_\Omega \nabla u \cdot \mathbf V \, d\Omega, }

or in other words Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_\Omega u\,\operatorname{div}(\mathbf V)\,d\Omega \ =\ \int_\Gamma u \mathbf V \cdot \hat{\mathbf n}\,d\Gamma - \int_\Omega \operatorname{grad}(u)\cdot\mathbf V\,d\Omega . } The regularity requirements of the theorem can be relaxed. For instance, the boundary Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Gamma=\partial\Omega} need only be Lipschitz continuous, and the functions u, v need only lie in the Sobolev space Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H^1(\Omega)} .

Green's first identity

Consider the continuously differentiable vector fields Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf U = u_1\mathbf e_1+\cdots+u_n\mathbf e_n} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v \mathbf e_1,\ldots, v\mathbf e_n} , where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf e_i} is the i-th standard basis vector for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i=1,\ldots,n} . Now apply the above integration by parts to each Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_i} times the vector field Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v\mathbf e_i} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_\Omega u_i\frac{\partial v}{\partial x_i}\,d\Omega \ =\ \int_\Gamma u_i v \,\mathbf e_i\cdot\hat\mathbf{n}\,d\Gamma - \int_\Omega \frac{\partial u_i}{\partial x_i} v\,d\Omega.}

Summing over i gives a new integration by parts formula:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_\Omega \mathbf U \cdot \nabla v\,d\Omega \ =\ \int_\Gamma v \mathbf{U}\cdot \hat{\mathbf n}\,d\Gamma - \int_\Omega v\, \nabla \cdot \mathbf{U}\,d\Omega.}

The case Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{U}=\nabla u} , where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u\in C^2(\bar{\Omega})} , is known as the first of Green's identities:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_\Omega \nabla u \cdot \nabla v\,d\Omega\ =\ \int_\Gamma v\, \nabla u\cdot\hat{\mathbf n}\,d\Gamma - \int_\Omega v\, \nabla^2 u \, d\Omega.}

Notes

^ "Brook Taylor". History.MCS.St-Andrews.ac.uk. Retrieved May 25, 2018.
^ "Brook Taylor". Stetson.edu. Archived from the original on January 3, 2018. Retrieved May 25, 2018.
^ "Integration by parts". Encyclopedia of Mathematics.
^ Kasube, Herbert E. (1983). "A Technique for Integration by Parts". The American Mathematical Monthly. 90 (3): 210–211. doi:10.2307/2975556. JSTOR 2975556.
^ Thomas, G. B.; Finney, R. L. (1988). Calculus and Analytic Geometry (7th ed.). Reading, MA: Addison-Wesley. ISBN 0-201-17069-8.
^ Horowitz, David (1990). "Tabular Integration by Parts" (PDF). The College Mathematics Journal. 21 (4): 307–311. doi:10.2307/2686368. JSTOR 2686368.
^ Rogers, Robert C. (September 29, 2011). "The Calculus of Several Variables" (PDF).

External links

"Integration by parts", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
Integration by parts—from MathWorld

es:Métodos de integración#Método de integración por partes

[Brook_Taylor_biography,_St._Andrews-1] "Brook Taylor". History.MCS.St-Andrews.ac.uk. Retrieved May 25, 2018.

[Brook_Taylor_biography,_Stetson-2] "Brook Taylor". Stetson.edu. Archived from the original on January 3, 2018. Retrieved May 25, 2018.

[3] "Integration by parts". Encyclopedia of Mathematics.

[4] Kasube, Herbert E. (1983). "A Technique for Integration by Parts". The American Mathematical Monthly. 90 (3): 210–211. doi:10.2307/2975556. JSTOR 2975556.

[5] Thomas, G. B.; Finney, R. L. (1988). Calculus and Analytic Geometry (7th ed.). Reading, MA: Addison-Wesley. ISBN 0-201-17069-8.

[6] Horowitz, David (1990). "Tabular Integration by Parts" (PDF). The College Mathematics Journal. 21 (4): 307–311. doi:10.2307/2686368. JSTOR 2686368.

[7] Rogers, Robert C. (September 29, 2011). "The Calculus of Several Variables" (PDF).

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v t e Integrals
Types of integrals	Riemann integral Lebesgue integral Burkill integral Bochner integral Daniell integral Darboux integral Henstock–Kurzweil integral Haar integral Hellinger integral Khinchin integral Kolmogorov integral Lebesgue–Stieltjes integral Pettis integral Pfeffer integral Riemann–Stieltjes integral Regulated integral
Integration techniques	Substitution Trigonometric Euler Weierstrass By parts Partial fractions Euler's formula Inverse functions Changing order Reduction formulas Parametric derivatives Differentiation under the integral sign Laplace transform Contour integration Laplace's method Numerical integration Simpson's rule Trapezoidal rule Risch algorithm
Improper integrals	Gaussian integral Dirichlet integral Fermi–Dirac integral complete incomplete Bose–Einstein integral Frullani integral Common integrals in quantum field theory
Stochastic integrals	Itô integral Russo–Vallois integral Stratonovich integral Skorokhod integral
Miscellaneous	Basel problem Euler–Maclaurin formula Gabriel's horn Integration Bee Proof that 22/7 exceeds π Volumes Washers Shells

Integration by parts

Theorem

Product of two functions

Validity for less smooth functions

Product of many functions

Visualization

Applications

Finding antiderivatives

Polynomials and trigonometric functions

Exponentials and trigonometric functions

Functions multiplied by unity

LIATE rule

Wallis product

Gamma function identity

Use in harmonic analysis

Fourier transform of derivative

Decay of Fourier transform

Use in operator theory

Other applications

Repeated integration by parts

Tabular integration by parts

Higher dimensions

Green's first identity

See also

Notes

Further reading

External links