Integration by Parts

Integration by parts is not confined to symbolic calculus exercises; it is a structural tool across analysis and its applications. In Fourier analysis, repeated integration by parts is the standard way to estimate the decay rate of Fourier coefficients of smooth functions, since each integration by parts trades a derivative on one factor for an extra power of $1/n$ in the coefficient bound. In probability and statistics, it underlies the derivation of moments of distributions via the identity relating $E[X^n]$ to integrals of the survival function, and appears in Stein's method for bounding distances between distributions. In partial differential equations and functional analysis, the formula generalizes to Green's identities and the divergence theorem, and directly motivates the definition of the weak derivative: a function $w$ is declared the weak derivative of $u$ precisely when it satisfies the integration-by-parts identity against all smooth test functions, even when $u$ itself is not classically differentiable. In quantum mechanics, integration by parts on the real line (with vanishing boundary terms at infinity) is how one shows the momentum operator is Hermitian. In numerical analysis and scientific computing, it is the mechanism behind deriving weak/variational formulations used in the finite element method. And in extending the definite-integral formula to unbounded domains — the natural next question after this article — one must carefully control the boundary term $u(b)v(b)$ as $b\to\infty$, which is precisely the technical heart of applying this identity within the theory of Improper Integrals.

Ask the tutor: How does integration by parts extend to improper integrals like $\int_0^\infty x e^{-x}\,dx$, and what exactly must be checked before taking the boundary term to infinity?

Think of the product rule as a statement about instantaneous rates: the rate of change of $u(x)v(x)$ splits into a piece where $u$ is held still and $v$ moves, plus a piece where $v$ is held still and $u$ moves. Integration by parts is what happens when you accumulate that splitting over an entire interval: the total change in the product $u(b)v(b)-u(a)v(a)$ decomposes into the total contribution of "$v$ moving against $u$" and the total contribution of "$u$ moving against $v$." If you know one of those two accumulated contributions, and you know the total change in the product, you get the other for free. Geometrically, picture the rectangle with corners at $(u(a), 0)$ and $(u(b), v(b))$ traced out as $x$ runs from $a$ to $b$, viewing $(u(x), v(x))$ as a curve in the plane. The area swept out relates to $\int u\,dv$ and $\int v\,du$ the way a rectangle's area splits into an "L-shaped" region under one curve and over the other — this is exactly the classical picture used to prove the formula for monotone functions without calculus, and it explains the name "parts": you cut one product-integral into two complementary pieces.

Example 1: $\displaystyle\int_0^{\pi/2} x\cos x\,dx$ (asymmetric algebraic-times-trigonometric). We want to apply the theorem with a good choice of which factor is $u$ and which is $v'$. Set

Check the hypotheses on $[a,b] = [0,\pi/2]$: $u(x)=x$ is continuous and differentiable everywhere with $u'(x)=1$, which is continuous, hence Riemann integrable, on $[0,\pi/2]$. Take $v(x) = \sin x$; then $v$ is continuous and differentiable with $v'(x)=\cos x$, continuous, hence Riemann integrable. All hypotheses of the theorem hold. By the theorem,

Compute the boundary term:

Compute the remaining integral using the Second Fundamental Theorem of Calculus with antiderivative $-\cos x$:

Therefore

Numerically, $\pi/2 \approx 1.5708$, so the value is approximately $0.5708$; this is consistent with the graph of $x\cos x$ on $[0,\pi/2]$, which is positive throughout the interval (since both $x\ge 0$ and $\cos x \ge 0$ there) but small near the right endpoint where $\cos x \to 0$, so a value near $0.57$ over an interval of length $\approx 1.5708$ is plausible ($0.5708/1.5708\approx 0.36$, a reasonable average height for a function starting at $0$, rising, then falling back toward $0$).

Example 2: $\displaystyle\int_1^{e} x^2\ln x\,dx$ (a case where the "wrong" choice would make things worse, shown explicitly, plus a double-check by direct antidifferentiation). Here the two natural choices are $u=\ln x, v' = x^2$ or $u = x^2, v' = \ln x$. We first show the bad choice is legal but unhelpful, then do the good choice. Bad choice check. Take $u(x) = x^2$, $v'(x) = \ln x$. On $[1,e]$, $u$ is continuous and differentiable, $u'(x) = 2x$ continuous. An antiderivative of $\ln x$ is $v(x) = x\ln x - x$ (verify: $v'(x) = \ln x + x\cdot\frac1x - 1 = \ln x$, correct), and $v$ is continuous and differentiable with $v'=\ln x$ continuous on $[1,e]$ (note $\ln x$ is continuous on $[1,e]$ since $1>0$, so there is no issue at $x=1$). All hypotheses hold, so the theorem applies, but it turns

into a harder integral on the right (still containing $x^2\ln x$ after expansion) — a legitimate application of the theorem that is simply a poor strategic choice. This illustrates that the theorem's hypotheses being satisfied does not guarantee usefulness; the choice of $u,v'$ is a matter of strategy, not validity. Good choice. Take $u(x) = \ln x$, $v'(x) = x^2$. Check hypotheses on $[1,e]$: $u(x)=\ln x$ is continuous and differentiable on $[1,e]$ (indeed on $(0,\infty)\supset[1,e]$) with $u'(x) = 1/x$, continuous on $[1,e]$ since $x\ge 1>0$ there, hence Riemann integrable. Take $v(x) = x^3/3$; $v$ is continuous and differentiable with $v'(x)=x^2$ continuous. All hypotheses of the theorem hold. By the theorem,

Evaluate the boundary term: at $x=e$, $\ln e = 1$, so this term is $e^3/3$; at $x=1$, $\ln 1 = 0$, so this term is $0$. Hence the boundary term is $e^3/3 - 0 = e^3/3$. Evaluate the remaining integral by the Second Fundamental Theorem of Calculus:

Therefore

Numerically, $e^3 \approx 20.0855$, so $2e^3 + 1 \approx 41.171$, and dividing by $9$ gives approximately $4.5746$. As a sanity check, $x^2\ln x$ ranges from $0$ at $x=1$ to $e^2 \approx 7.389$ at $x=e$ over an interval of length $e-1\approx 1.718$; an average value near $4.57/1.718 \approx 2.66$ is plausible for a function increasing from $0$ to about $7.39$, since it is convex-looking and weighted toward the larger values near $x=e$.