Integration by parts

If, \(h(x)=f(x)g(x),\) then by using the product rule, we obtain \({h}^{'}(x)={f}^{'}(x)g(x)+{g}^{'}(x)f(x).\) Although at first it may seem counterproductive, let’s now integrate both sides of this equation: \(\int {h}^{'}(x)dx=\int (g(x){f}^{'}(x)+f(x){g}^{'}(x))dx.\)

This gives us

Now we solve for \(\int f(x){g}^{'}(x)dx:\)

By making the substitutions \(u=f(x)\) and \(v=g(x),\) which in turn make \(du={f}^{'}(x)dx\) and \(dv={g}^{'}(x)dx,\) we have the more compact form

The advantage of using the integration-by-parts formula is that we can use it to exchange one integral for another, possibly easier, integral. The following example illustrates its use.

The natural question to ask at this point is: How do we know how to choose \(u\) and \(dv?\) Sometimes it is a matter of trial and error; however, the acronym LIATE can often help to take some of the guesswork out of our choices. This acronym stands for Logarithmic Functions, Inverse Trigonometric Functions, Algebraic Functions, Trigonometric Functions, and Exponential Functions. This mnemonic serves as an aid in determining an appropriate choice for \(u.\)

Condensed: the full section is in OpenStax Calculus Volume 2.

Now that we have used integration by parts successfully to evaluate indefinite integrals, we turn our attention to definite integrals. The integration technique is really the same, only we add a step to evaluate the integral at the upper and lower limits of integration.

• The integration-by-parts formula allows the exchange of one integral for another, possibly easier, integral.
• Integration by parts applies to both definite and indefinite integrals.

In using the technique of integration by parts, you must carefully choose which expression is u. For each of the following problems, use the guidelines in this section to choose u. Do not evaluate the integrals.

Find the integral by using the simplest method. Not all problems require integration by parts.

Compute the definite integrals. Use a graphing utility to confirm your answers.

Derive the following formulas using the technique of integration by parts. Assume that n is a positive integer. These formulas are called reduction formulas because the exponent in the x term has been reduced by one in each case. The second integral is simpler than the original integral.

State whether you would use integration by parts to evaluate the integral. If so, identify u and dv. If not, describe the technique used to perform the integration without actually doing the problem.

Sketch the region bounded above by the curve, the x-axis, and \(x=1,\) and find the area of the region. Provide the exact form or round answers to the number of places indicated.

Find the volume generated by rotating the region bounded by the given curves about the specified line. Express the answers in exact form or approximate to the number of decimal places indicated.

In Section, we learned the technique of \(u\)-substitution for evaluating indefinite integrals. For example, the indefinite integral \(\int x^3 \sin(x^4) \, dx\) is perfectly suited to \(u\)-substitution, because one factor is a composite function and the other factor is the derivative (up to a constant) of the inner function. Recognizing the algebraic structure of a function can help us to find its antiderivative.

Next we consider integrands with a different elementary algebraic structure: a product of basic functions. For instance, suppose we are interested in evaluating the indefinite integral \[\begin{aligned}\end{aligned}\].

The integrand is the product of the basic functions \(f(x) = x\) and \(g(x) = \sin(x)\). We know that it is relatively complicated to compute the derivative of the product of two functions, so we should expect that antidifferentiating a product should be similarly involved. Intuitively, we expect that evaluating \(\int x \sin(x) \, dx\) will involve somehow reversing the Product Rule.

To that end, in Preview Activity we refresh our understanding of the Product Rule and then investigate some indefinite integrals that involve products of basic functions.

Problem (c) in Preview Activity provides a clue to the general technique known as Integration by Parts, which comes from reversing the Product Rule. Recall that the Product Rule states that \[\begin{aligned}\end{aligned}\].

Integrating both sides of this equation indefinitely with respect to \(x\), we find \[\begin{aligned}\end{aligned}\].

On the left side of Equation, we have the indefinite integral of the derivative of a function. Temporarily omitting the constant that may arise, we have \[\begin{aligned}\end{aligned}\].

We solve for the first indefinite integral on the left to generate the rule \[\begin{aligned}\end{aligned}\].

Often we express Equation in terms of the variables \(u\) and \(v\), where \(u = f(x)\) and \(v = g(x)\). In differential notation, \(du = f'(x) \, dx\) and \(dv = g'(x) \, dx\), so we can state the rule for Integration by Parts in its most common form as follows:

\[\begin{aligned}\end{aligned}\].

To apply integration by parts, we look for a product of basic functions that we can identify as \(u\) and \(dv\). If we can antidifferentiate \(dv\) to find \(v\), and evaluating \(\int v \, du\) is not more difficult than evaluating \(\int u \, dv\), then this substitution usually proves to be fruitful. To demonstrate, we consider the following example.

The general technique of integration by parts involves trading the problem of integrating the product of two functions for the problem of integrating the product of two related functions. That is, we convert the problem of evaluating \(\int u \, dv\) to that of evaluating \(\int v \, du\). This clearly shapes our choice of \(u\) and \(v\). In Example, the original integral to evaluate was \(\int x \cos(x) \,dx\), and through the substitution provided by integration by parts, we were instead able to evaluate \(\int \sin(x) \cdot 1 \, dx\). Note that the original function \(x\) was replaced by its derivative, while \(\cos(x)\) was replaced by its antiderivative.

Condensed: the full section is in Boelkins, Active Calculus.

Sometimes integration by parts is not an obvious choice, but the technique is appropriate nonetheless. Integration by parts allows us to replace one function in a product with its derivative while replacing the other with its antiderivative. For instance, consider evaluating \[\begin{aligned}\end{aligned}\].

Initially, this problem seems ill-suited to integration by parts, since there does not appear to be a product of functions present. But if we note that \(\arctan(x) = \arctan(x) \cdot 1\), and realize that we know the derivative of \(\arctan(x)\) as well as the antiderivative of \(1\), we see the possibility for the substitution \(u = \arctan(x)\) and \(dv = 1 \, dx\). We explore this substitution further in Activity.

In a related problem, consider \(\int t^3 \sin(t^2) \, dt\). Observe that there is a composite function present in \(\sin(t^2)\), but there is not an obvious function-derivative pair, as we have \(t^3\) (rather than simply \(t\)) multiplying \(\sin(t^2)\). In this problem we use both \(u\)-substitution and integration by parts. First we write \(t^3 = t \cdot t^2\) and consider the indefinite integral \[\begin{aligned}\end{aligned}\]. We let \(z = t^2\) so that \(dz = 2t \, dt\), and thus \(t \, dt = \frac{1}{2} \, dz\). (We are using the variable \(z\) to perform a \(z\)-substitution first so that we may then apply integration by parts.) Under this \(z\)-substitution, we now have \[\begin{aligned}\end{aligned}\].

The resulting integral can be evaluated by parts. This, too, is explored further in Activity.

These problems show that we sometimes must think creatively in choosing the variables for substitution in integration by parts, and that we may need to use substitution for an additional change of variables.

Integration by parts is well suited to integrating the product of basic functions, allowing us to trade a given integrand for a new one where one function in the product is replaced by its derivative, and the other is replaced by its antiderivative. The goal in this trade of \(\int u \, dv\) for \(\int v \, du\) is that the new integral be simpler to evaluate than the original one. Sometimes it is necessary to apply integration by parts more than once in order to evaluate a given integral.

Of course, even more than two applications of integration by parts may be necessary. In the preceding example, if the integrand had been \(t^3e^t\), we would have had to use integration by parts three times.

Next, we consider the slightly different scenario.

Condensed: the full section is in Boelkins, Active Calculus.

We can use the technique of integration by parts to evaluate a definite integral as well.

As with any substitution technique, it is important to use notation carefully and completely, and to ensure that the end result makes sense.

Both integration techniques we have discussed apply in relatively limited circumstances. It is not hard to find examples of functions for which neither technique produces an antiderivative; indeed, there are many, many functions that appear elementary but that do not have an elementary algebraic antiderivative. For instance, neither \(u\)-substitution nor integration by parts proves fruitful for the indefinite integrals \[\begin{aligned}\end{aligned}\]. While there are other integration techniques, some of which we will consider briefly, none of them enables us to find an algebraic antiderivative for \(e^{x^2}\) or \(x \tan(x)\). We do know from the Second Fundamental Theorem of Calculus that we can construct an integral antiderivative for each function; \(F(x) = \int_0^x e^{t^2} \, dt\) is an antiderivative of \(f(x) = e^{x^2}\), and \(G(x) = \int_0^{x} t \tan(t) \, dt\) is an antiderivative of \(g(x) = x \tan(x)\). But finding an elementary algebraic formula that doesn't involve integrals for either \(F\) or \(G\) turns out not only to be impossible through \(u\)-substitution or integration by parts, but indeed impossible altogether. Antidifferentiation is much harder in general than differentiation.

• Through the method of integration by parts, we can evaluate indefinite integrals that involve products of basic functions such as \(\int x \sin(x) \, dx\) and \(\int x \ln(x) \, dx\). Using a substitution enables us to trade one of the functions in the product for its derivative, and the other for its antiderivative, in an effort to find a different product of functions that is easier to integrate.

• If the algebraic structure of an integrand is a product of basic functions in the form \(\int f(x) g'(x) \, dx\), we can use the substitution \(u = f(x)\) and \(dv = g'(x) \,dx\) and apply the rule \[\begin{aligned}\end{aligned}\] to evaluate the original integral \(\int f(x) g'(x) \, dx\) by instead evaluating \[\begin{aligned}\end{aligned}\].

• When deciding to integrate by parts, we have to select both \(u\) and \(dv\). That selection is guided by the overall principle that the new integral \(\int v \, du\) not be more difficult than the original integral \(\int u \, dv\). In addition, it is often helpful to recognize if one of the functions present is much easier to differentiate than antidifferentiate (such as \(\ln(x)\)), in which case that function often is best assigned the variable \(u\). In addition, \(dv\) must be a function that we can antidifferentiate.

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 \(v = v(x)\) and \(dv = v'(x) \, dx,\) the formula can be written more compactly: \[\int u \, dv \ =\ uv - \int v \, du.\]

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. 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.

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 \(x\),

\[\int \Big(u(x)v(x)\Big)'\,dx = \int u'(x)v(x)\,dx + \int u(x)v'(x) \,dx,\]

and noting that an indefinite integral is an antiderivative gives

\[u(x)v(x) = \int u'(x)v(x)\,dx + \int u(x)v'(x)\,dx,\]

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

Condensed: the full section is in Wikipedia.

It is not necessary for \(u\) and \(v\) to be continuously differentiable. Integration by parts works if \(u\) is absolutely continuous and the function designated \(v'\) is Lebesgue integrable (but not necessarily continuous). (If \(v'\) has a point of discontinuity then its antiderivative \(v\) may not have a derivative at that point.)

If the interval of integration is not compact, then it is not necessary for \(u\) to be absolutely continuous in the whole interval or for \(v'\) to be Lebesgue integrable in the interval, as a couple of examples (in which \(u\) and \(v\) are continuous and continuously differentiable) will show. For instance, if

\[u(x)= e^x/x^2, \, v'(x) =e^{-x}\]

\(u\) is not absolutely continuous on the interval [1, ∞), but nevertheless:

\[\int_1^\infty u(x)v'(x)\,dx = \Big[u(x)v(x)\Big]_1^\infty - \int_1^\infty u'(x)v(x)\,dx\]

so long as \(\left[u(x)v(x)\right]_1^\infty\) is taken to mean the limit of \(u(L)v(L)-u(1)v(1)\) as \(L\to\infty\) and so long as the two terms on the right-hand side are finite. This is only true if we choose \(v(x)=-e^{-x}.\) Similarly, if

\[u(x)= e^{-x},\, v'(x) =x^{-1}\sin(x)\]

Condensed: the full section is in Wikipedia.

Integrating the product rule for three multiplied functions, \(u(x)\), \(v(x)\), \(w(x)\), gives a similar result:

\[\int_a^b u v \, dw \ =\ \Big[u v w\Big]^b_a - \int_a^b u w \, dv - \int_a^b v w \, du.\]

In general, for \(n\) factors

\[\left(\prod_{i=1}^n u_i(x) \right)' \ =\ \sum_{j=1}^n u_j'(x)\prod_{i\neq j}^n u_i(x),\]

which leads to

\[\left[ \prod_{i=1}^n u_i(x) \right]_a^b \ =\ \sum_{j=1}^n \int_a^b u_j'(x) \prod_{i\neq j}^n u_i(x).\]

Consider a parametric curve \((x, y) = (f(t), g(t))\). Assuming that the curve is locally one-to-one and integrable, we can define \[\begin{aligned} x(y) &= f(g^{-1}(y)) \\ y(x) &= g(f^{-1}(x)) \end{aligned}\]

The area of the blue region is

\[A_1=\int_{y_1}^{y_2}x(y) \, dy\]

Similarly, the area of the red region is \[A_2=\int_{x_1}^{x_2}y(x)\,dx\]

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

\[\overbrace{\int_{y_1}^{y_2}x(y) \, dy}^{A_1}+\overbrace{\int_{x_1}^{x_2}y(x) \, dx}^{A_2}\ =\ \overbrace{x(y_2) \cdot y(x_2)}^{final} - \overbrace{x(y_1) \cdot y(x_1)}^{initial} =\ \biggl.x \cdot y(x)\biggl|_{x_1}^{x_2} \ =\ \biggl.y \cdot x(y)\biggl|_{y_1}^{y_2}\] 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 \[\int x\,dy + \int y \,dx \ =\ xy\] Rearranging: \[\int x\,dy \ =\ xy - \int y \,dx\] 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(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 \(f\) 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 \(f^{-1}\) in terms of the integral of \(f\). This is demonstrated in the article, Integral of inverse functions.

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:

\[\int uv\,dx = u \int v\,dx - \int\left(u' \int v\,dx \right)\,dx.\]

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:

\[\int\frac{\ln(x)}{x^2}\,dx\,.\]

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

\[\int\frac{\ln(x)}{x^2}\,dx = -\frac{\ln(x)}{x} - \int \biggl(\frac1{x}\biggr) \biggl(-\frac1{x}\biggr)\,dx\,.\]

The antiderivative of −⁠1/x⁠ can be found with the power rule and is ⁠1/x⁠, making the final integral

Condensed: the full section is in Wikipedia.

The Wallis infinite product for \(\pi\)

\[\begin{aligned} \frac{\pi}{2} & = \prod_{n=1}^\infty \frac{ 4n^2 }{ 4n^2 - 1 } = \prod_{n=1}^\infty \left(\frac{2n}{2n-1} \cdot \frac{2n}{2n+1}\right) \\[6pt] & = \Big(\frac{2}{1} \cdot \frac{2}{3}\Big) \cdot \Big(\frac{4}{3} \cdot \frac{4}{5}\Big) \cdot \Big(\frac{6}{5} \cdot \frac{6}{7}\Big) \cdot \Big(\frac{8}{7} \cdot \frac{8}{9}\Big) \cdot \; \cdots \end{aligned}\]

may be derived using integration by parts.

The gamma function is an example of a special function, defined as an improper integral for \(z > 0\). Integration by parts illustrates it to be an extension of the factorial function:

\[\begin{aligned} \Gamma(z) & = \int_0^\infty e^{-x} x^{z-1} dx \\[6pt] & = - \int_0^\infty x^{z-1} \, d\left(e^{-x}\right) \\[6pt] & = - \Biggl[e^{-x} x^{z-1}\Biggl]_0^\infty + \int_0^\infty e^{-x} d\left(x^{z-1}\right) \\[6pt] & = 0 + \int_0^\infty \left(z-1\right) x^{z-2} e^{-x} dx\\[6pt] & = (z-1)\Gamma(z-1). \end{aligned}\]

Since

\[\Gamma(1) = \int_0^\infty e^{-x} \, dx = 1,\]

when \(z\) is a natural number, that is, \(z = n \in \mathbb{N}\), applying this formula repeatedly gives the factorial: \(\Gamma(n+1) = n!\)

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.

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 \(L^2\) (see L space). If \(f\) is smooth and compactly supported then, using integration by parts, we have

\[\begin{aligned} \langle -\Delta f, f \rangle_{L^2} &= -\int_{-\infty}^\infty f''(x)\overline{f(x)}\,dx \\[5pt] &=-\left[f'(x)\overline{f(x)}\right]_{-\infty}^\infty + \int_{-\infty}^\infty f'(x)\overline{f'(x)}\,dx \\[5pt] &=\int_{-\infty}^\infty \vert f'(x)\vert^2\,dx \geq 0. \end{aligned}\]

Considering a second derivative of \(v\) in the integral on the LHS of the formula for partial integration suggests a repeated application to the integral on the RHS: \[\int u v''\,dx = uv' - \int u'v'\,dx = uv' - \left( u'v - \int u''v\,dx \right).\]

Extending this concept of repeated partial integration to derivatives of degree n leads to \[\begin{aligned} \int u^{(0)} v^{(n)}\,dx &= u^{(0)} v^{(n-1)} - u^{(1)}v^{(n-2)} + u^{(2)}v^{(n-3)} - \cdots + (-1)^{n-1}u^{(n-1)} v^{(0)} + (-1)^n \int u^{(n)} v^{(0)} \,dx.\\[5pt] &= \sum_{k=0}^{n-1}(-1)^k u^{(k)}v^{(n-1-k)} + (-1)^n \int u^{(n)} v^{(0)} \,dx. \end{aligned}\]

This concept may be useful when the iterated integrals of \(v^{(n)}\) are readily available (e.g., plain exponentials or sine and cosine, as in Laplace or Fourier transforms), and when the nth derivative of \(u\) vanishes (e.g., as a polynomial function with degree \((n-1)\)). 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 \[\int u^{(0)} v^{(n)}\,dx \quad\] and \(\quad \int u^{(\ell)} v^{(n-\ell)}\,dx \quad\) and \(\quad \int u^{(m)} v^{(n-m)}\,dx \quad\text{ for } 1 \le m,\ell \le n\) get related. This may be interpreted as arbitrarily "shifting" derivatives between \(v\) and \(u\) within the integrand, and proves useful, too (see Rodrigues' formula).

The essential process of the above formula can be summarized in a table; the resulting method is called "tabular integration".

For example, consider the integral

\[\int x^3 \cos x \,dx \quad\] and take \(\quad u^{(0)} = x^3, \quad v^{(n)} = \cos x.\)

Begin to list in column A the function \(u^{(0)} = x^3\) and its subsequent derivatives \(u^{(i)}\) until zero is reached. Then list in column B the function \(v^{(n)} = \cos x\) and its subsequent integrals \(v^{(n-i)}\) until the size of column B is the same as that of column A. The result is as follows:

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 ith integral must be added to all the previous products (0 ≤ j < i) of the jth 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 jth 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):

\[\underbrace{(+1)(x^3)(\sin x)}_{j=0} + \underbrace{(-1)(3x^2)(-\cos x)}_{j=1} + \underbrace{(+1)(6x)(-\sin x)}_{j=2} +\underbrace{(-1)(6)(\cos x)}_{j=3}+ \underbrace{\int(+1)(0)(\cos x) \,dx}_{i=4: \;\to \;C}.\]

This yields

Condensed: the full section is in Wikipedia.

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.

The product rule for divergence states:

\[\nabla \cdot ( u \mathbf{V} ) \ =\ u\, \nabla \cdot \mathbf V \ +\ \nabla u\cdot \mathbf V.\]

Suppose \(\Omega\) is an open bounded subset of \(\R^n\) with a piecewise smooth boundary \(\Gamma=\partial\Omega\). Integrating over \(\Omega\) with respect to the standard volume form \(d\Omega\), and applying the divergence theorem, gives:

\[\int_{\Gamma} u \mathbf{V} \cdot \hat{\mathbf n} \,d\Gamma \ =\ \int_\Omega\nabla\cdot ( u \mathbf{V} )\,d\Omega \ =\ \int_\Omega u\, \nabla \cdot \mathbf V\,d\Omega \ +\ \int_\Omega\nabla u\cdot \mathbf V\,d\Omega,\]

where \(\hat{\mathbf n}\) is the outward unit normal vector to the boundary, integrated with respect to its standard Riemannian volume form \(d\Gamma\). Rearranging gives:

\[\int_\Omega u \,\nabla \cdot \mathbf V\,d\Omega \ =\ \int_\Gamma u \mathbf V \cdot \hat{\mathbf n}\,d\Gamma - \int_\Omega \nabla u \cdot \mathbf V \, d\Omega,\]

Condensed: the full section is in Wikipedia.

Consider the continuously differentiable vector fields \(\mathbf U = u_1\mathbf e_1+\cdots+u_n\mathbf e_n\) and \(v \mathbf e_1,\ldots, v\mathbf e_n\), where \(\mathbf e_i\) is the i-th standard basis vector for \(i=1,\ldots,n\). Now apply the above integration by parts to each \(u_i\) times the vector field \(v\mathbf e_i\):

\[\int_\Omega u_i\frac{\partial v}{\partial x_i}\,d\Omega \ =\ \int_\Gamma u_i v \,\mathbf e_i\cdot\hat{\mathbf{n}}\,d\Gamma - \int_\Omega \frac{\partial u_i}{\partial x_i} v\,d\Omega.\]

Summing over i gives a new integration by parts formula:

\[\int_\Omega \mathbf U \cdot \nabla v\,d\Omega \ =\ \int_\Gamma v \mathbf{U}\cdot \hat{\mathbf n}\,d\Gamma - \int_\Omega v\, \nabla \cdot \mathbf{U}\,d\Omega.\]

The case \(\mathbf{U}=\nabla u\), where \(u\in C^2(\bar{\Omega})\), is known as the first of Green's identities:

\[\int_\Omega \nabla u \cdot \nabla v\,d\Omega\ =\ \int_\Gamma v\, \nabla u\cdot\hat{\mathbf n}\,d\Gamma - \int_\Omega v\, \nabla^2 u \, d\Omega.\]