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s 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 x 2 2 cos x x 2 2 sin x d x displaystyle 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 x 3 e x 2 d x displaystyle int x 3 e x 2 dx one would set u x 2 d v x e x 2 d x displaystyle u x 2 quad dv x cdot e x 2 dx so that d u 2 x d x v e x 2 2 displaystyle du 2x dx quad v frac e x 2 2 then x 3 e x 2 d x x 2 x e x 2 d x u d v u v v d u x 2 e x 2 2 x e x 2 d x 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 xe x 2 dx finally this results in x 3 e x 2 d x e x 2 x 2 1 2 c 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 edit the wallis infinite product for π displaystyle pi π 2 n 1 4 n 2 4 n 2 1 n 1 2 n 2 n 1 2 n 2 n 1 2 1 2 3 4 3 4 5 6 5 6 7 8 7 8 9 displaystyle 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 gamma function identity edit the gamma function is an example of a special function defined as an improper integral for z 0 displaystyle z 0 integration by parts illustrates it to be an extension of the factorial function γ z 0 e x x z 1 d x 0 x z 1 d e x e x x z 1 0 0 e x d x z 1 0 0 z 1 x z 2 e x d x z 1 γ z 1 displaystyle 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 γ 1 0 e x d x 1 displaystyle gamma 1 int _ 0 infty e x dx 1 when z displaystyle z is a natural number that is z n n displaystyle z n in mathbb n applying this formula repeatedly gives the factorial γ n 1 n displaystyle gamma n 1 n use in harmonic analysis edit 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 edit if f displaystyle f is a k displaystyle k times continuously differentiable function and all derivatives up to the k displaystyle k th one decay to zero at infinity then its fourier transform satisfies f f k ξ 2 π i ξ k f f ξ displaystyle mathcal f f k xi 2 pi i xi k mathcal f f xi where f k displaystyle f k is the k displaystyle k th derivative of f displaystyle f the exact constant on the right depends on the convention of the fourier transform used this is proved by noting that d d y e 2 π i y ξ 2 π i ξ e 2 π i y ξ 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 f f ξ e 2 π i y ξ f y d y e 2 π i y ξ f y 2 π i ξ e 2 π i y ξ f y d y 2 π i ξ e 2 π i y ξ f y d y 2 π i ξ f f ξ displaystyle begin aligned 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 aligned applying this inductively gives the result for general k displaystyle k a similar method can be used to find the laplace transform of a derivative of a function decay of fourier transform edit the above result tells us about the decay of the fourier transform since it follows that if f displaystyle f and f k displaystyle f k are integrable then f f ξ i f 1 2 π ξ k where i f f y f k y d y 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 f displaystyle f satisfies these conditions then its fourier transform decays at infinity at least as quickly as 1 ξ k in particular if k 2 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 f f ξ f y d y 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 2 π i ξ k f f ξ f k y d y 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 edit 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 displaystyle l 2 see l p space if f displaystyle f is smooth and compactly supported then using integration by parts we have δ f f l 2 f x f x d x f x f x f x f x d x f x 2 d x 0 displaystyle 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 other applications edit determining boundary conditions in sturm liouville theory deriving the euler lagrange equation in the calculus of variations repeated integration by parts edit further information cauchy formula for repeated integration considering a second derivative of v displaystyle v in the integral on the lhs of the formula for partial integration suggests a repeated application to the integral on the rhs u v d x u v u v d x u v u v u v d x displaystyle int uv 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 u 0 v n d x u 0 v n 1 u 1 v n 2 u 2 v n 3 1 n 1 u n 1 v 0 1 n u n v 0 d x k 0 n 1 1 k u k v n 1 k 1 n u n v 0 d x displaystyle 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 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 u displaystyle u vanishes e g as a polynomial function with degree n 1 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 u 0 v n d x displaystyle int u 0 v n dx quad and u ℓ v n ℓ d x displaystyle quad int u ell v n ell dx quad and u m v n m d x for 1 m ℓ n displaystyle quad int u m v n m dx quad text for 1 leq m ell leq n get related this may be interpreted as arbitrarily shifting derivatives between v displaystyle v and u displaystyle u within the integrand and proves useful too see rodrigues formula tabular integration by parts edit the essential process of the above formula can be summarized in a table the resulting method is called tabular integration 5 for example consider the integral x 3 cos x d x displaystyle int x 3 cos x dx quad and take u 0 x 3 v n cos x displaystyle quad u 0 x 3 quad v n cos x begin to list in column a the function u 0 x 3 displaystyle u 0 x 3 and its subsequent derivatives u i displaystyle u i until zero is reached then list in column b the function v n cos x displaystyle v n cos x and its subsequent integrals v n i 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 u i displaystyle u i b integrals v n i displaystyle v n i 0 x 3 displaystyle x 3 cos x displaystyle cos x 1 3 x 2 displaystyle 3x 2 sin x displaystyle sin x 2 6 x displaystyle 6x cos x displaystyle cos x 3 6 displaystyle 6 sin x displaystyle sin x 4 0 displaystyle 0 cos x 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 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 1 x 3 sin x j 0 1 3 x 2 cos x j 1 1 6 x sin x j 2 1 6 cos x j 3 1 0 cos x d x i 4 c 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 x 3 cos x d x step 0 x 3 sin x 3 x 2 cos x 6 x sin x 6 cos x c 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 u i displaystyle u i and v n i 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 e x cos x d x displaystyle int e x cos x dx i sign a derivatives u i displaystyle u i b integrals v n i displaystyle v n i 0 e x displaystyle e x cos x displaystyle cos x 1 e x displaystyle e x sin x displaystyle sin x 2 e x displaystyle e x cos x 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 e x cos x d x step 0 1 e x sin x j 0 1 e x cos x j 1 1 e x cos x d x i 2 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 c displaystyle c and bringing the abstract integral to the other side gives 2 e x cos x d x e x sin x e x cos x c displaystyle 2 int e x cos x dx e x sin x e x cos x c and finally e x cos x d x 1 2 e x sin x cos x c displaystyle int e x cos x dx frac 1 2 left e x sin x cos x right c where c c 2 displaystyle c frac c 2 higher dimensions edit 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 6 the product rule for divergence states u v u v u v displaystyle nabla cdot u mathbf v u nabla cdot mathbf v nabla u cdot mathbf v suppose ω displaystyle omega is an open bounded subset of r n displaystyle mathbb r n with a piecewise smooth boundary γ ω displaystyle gamma partial omega integrating over ω displaystyle omega with respect to the standard volume form d ω displaystyle d omega and applying the divergence theorem gives γ u v n d γ ω u v d ω ω u v d ω ω u v d ω 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 n displaystyle hat mathbf n is the outward unit normal vector to the boundary integrated with respect to its standard riemannian volume form d γ displaystyle d gamma rearranging gives ω u v d ω γ u v n d γ ω u v d ω 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 ω u div v d ω γ u v n d γ ω grad u v d ω 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 γ ω displaystyle gamma partial omega need only be lipschitz continuous and the functions u v need only lie in the sobolev space h 1 ω displaystyle h 1 omega green s first identity edit consider the continuously differentiable vector fields u u 1 e 1 u n e n displaystyle mathbf u u_ 1 mathbf e _ 1 cdots u_ n mathbf e _ n and v e 1 v e n displaystyle v mathbf e _ 1 ldots v mathbf e _ n where e i displaystyle mathbf e _ i is the i th standard basis vector for i 1 n displaystyle i 1 ldots n now apply the above integration by parts to each u i displaystyle u_ i times the vector field v e i displaystyle v mathbf e _ i ω u i v x i d ω γ u i v e i n d γ ω u i x i v d ω 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 ω u v d ω γ v u n d γ ω v u d ω 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 u u displaystyle mathbf u nabla u where u c 2 ω displaystyle u in c 2 bar omega is known as the first of green s identities ω u v d ω γ v u n d γ ω v 2 u d ω 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 differential forms edit the exterior derivative on differential forms satisfies a graded product rule if u ω k m displaystyle u in omega k m is a differential k displaystyle k form and v ω n m displaystyle v in omega n m a differential n displaystyle n form on an orientable k n 1 displaystyle k n 1 manifold with boundary m displaystyle m then the graded product rule is d u v d u v 1 k u d v displaystyle d u wedge v du wedge v 1 k u wedge dv integrating both sides of the equality over m displaystyle m gives m d u v m d u v 1 k m u d v displaystyle int limits _ m d u wedge v int limits _ m du wedge v 1 k int limits _ m u wedge dv this is equivalent to m d u v m d u v 1 k m u d v displaystyle int limits _ m du wedge v int limits _ m d u wedge v 1 k int limits _ m u wedge dv an application of generalized stokes...