Worked Integration by Parts With Lebesgue Integrals

Using Integration by Parts in Lebesgue Integrals

Integration by parts is a practical method for transferring a derivative from one factor to another. In Lebesgue integration, the formula is justified by absolute continuity on a finite interval; on an unbounded interval, it is applied first on finite intervals and then a limit is taken. That last step requires checking both integrability and the boundary term. The examples below make these checks part of the calculation.

Unless stated otherwise, $m$ denotes Lebesgue measure on $\mathbb{R}$, and integrals over intervals use its restriction. The functions in the examples are real-valued. A function $u:[a,b]\to\mathbb{R}$ is absolutely continuous if, for every $\varepsilon>0$, there is a $\delta>0$ such that for every finite disjoint family of intervals $(a_j,b_j)\subseteq[a,b]$ with $\sum_j(b_j-a_j)<\delta$, we have $\sum_j|u(b_j)-u(a_j)|<\varepsilon$. Such a function has a derivative $u'$ $m$-a.e., with $u'\in L^1([a,b],m)$, and satisfies $u(y)-u(x)=\int_x^y u'(t)\,dt$ for every $x,y\in[a,b]$.

Integration-by-parts formula: If $u,v:[a,b]\to\mathbb{R}$ are absolutely continuous, then $uv$ is absolutely continuous and $$ \int_a^b u(x)v'(x)\,dx = u(b)v(b)-u(a)v(a)-\int_a^b u'(x)v(x)\,dx. $$ Both integrals are finite Lebesgue integrals. The derivatives are defined a.e.; changing their values on a null set does not affect the integrals. This is the integration-by-parts theorem for absolutely continuous functions.

The finite-interval formula is the starting point, not the conclusion, for an integral over $[0,\infty)$. We need to know that the integrands are integrable and that the product at the upper endpoint has a limit. The following elementary estimate supplies both checks for the examples involving integer powers.

Lemma: Let $\lambda>0$ and let $k$ be a nonnegative integer. On $(0,\infty)$, $x^k e^{-\lambda x}$ is integrable and $x^k e^{-\lambda x}\to0$ as $x\to\infty$.

Proof. For $x\ge0$, the exponential series gives $e^{\lambda x/2}\ge(\lambda x/2)^k/k!$ when $k\ge1$, and $e^{\lambda x/2}\ge1$ when $k=0$. Consequently, $$ 0\le x^k e^{-\lambda x} \le k!(2/\lambda)^k e^{-\lambda x/2}, $$ where for $k=0$ the constant on the right is $1$. The right-hand side is integrable on $(0,\infty)$, since its integral is $k!(2/\lambda)^k(2/\lambda)$. This proves integrability by comparison. The same bound tends to zero as $x\to\infty$, proving the limit. $\square$

Exponential Times an Integer Power

For $k\ge1$, apply integration by parts on $[0,R]$ with $u(x)=x^k$ and $v'(x)=e^{-\lambda x}$, so $v(x)=-e^{-\lambda x}/\lambda$. The finite-interval identity is $$ \int_0^R x^k e^{-\lambda x}\,dx = -\frac{R^k e^{-\lambda R}}{\lambda} +\frac{k}{\lambda}\int_0^R x^{k-1}e^{-\lambda x}\,dx. $$ The lower boundary term is zero because $k\ge1$. The lemma shows that the upper boundary term tends to zero and that both integrands are integrable on $(0,\infty)$. Taking $R\to\infty$ therefore gives a valid recurrence, rather than a formal integration by parts at infinity.

Recurrence: For every integer $k\ge1$ and every $\lambda>0$, $$ \int_0^\infty x^k e^{-\lambda x}\,dx = \frac{k}{\lambda}\int_0^\infty x^{k-1}e^{-\lambda x}\,dx. $$ Since $\int_0^\infty e^{-\lambda x}\,dx=1/\lambda$, repeated application gives $\int_0^\infty x^k e^{-\lambda x}\,dx=k!/\lambda^{k+1}$.

Worked Example: Computing an Exponential Times a Cubic

Compute $\int_0^\infty x^3e^{-2x}\,dx$. For each $k=0,1,2,3$, the lemma applies with $\lambda=2$, so $x^ke^{-2x}$ is integrable and $R^ke^{-2R}\to0$. The base integral is $$ \int_0^\infty e^{-2x}\,dx = \lim_{R\to\infty}\left[-\frac12e^{-2x}\right]_0^R = \frac12. $$ The recurrence then gives, with each step evaluated, $$ \int_0^\infty xe^{-2x}\,dx =\frac12\cdot\frac12=\frac14,\qquad \int_0^\infty x^2e^{-2x}\,dx =\frac22\cdot\frac14=\frac14, $$ $$ \int_0^\infty x^3e^{-2x}\,dx =\frac32\cdot\frac14=\frac38. $$ Check: an antiderivative of $x^3e^{-2x}$ is $-e^{-2x}(x^3/2+3x^2/4+3x/4+3/8)$. Its value at $0$ is $-3/8$, and its limit at infinity is $0$ by the lemma. The definite integral is therefore $0-(-3/8)=3/8$, as above.

Worked Example: A Different Power and Exponential Rate

Compute $\int_0^\infty x^4e^{-3x}\,dx$. The lemma with $\lambda=3$ verifies the integrability of each power $x^ke^{-3x}$, $0\le k\le4$, and the vanishing of each required boundary term. Starting with $\int_0^\infty e^{-3x}\,dx=1/3$, the recurrence gives $$ \int_0^\infty xe^{-3x}\,dx=\frac13\cdot\frac13=\frac19,\quad \int_0^\infty x^2e^{-3x}\,dx=\frac23\cdot\frac19=\frac{2}{27}, $$ $$ \int_0^\infty x^3e^{-3x}\,dx=\frac33\cdot\frac{2}{27}=\frac{2}{27},\quad \int_0^\infty x^4e^{-3x}\,dx=\frac43\cdot\frac{2}{27}=\frac{8}{81}. $$ Check: the recurrence’s closed expression gives $4!/3^5=24/243=8/81$. The denominator $3^5$ has one power from the base integral and one additional power for each of the four recurrence steps.

Piecewise Absolutely Continuous Factors

A factor need not be differentiable at every point to be absolutely continuous. In particular, a continuous function that is absolutely continuous on each interval of a finite partition is absolutely continuous on the whole interval. Integration by parts can be applied on each piece; the boundary terms at a join cancel because the factors are continuous there.

Proposition: Let $a=t_0<t_1<\cdots<t_N=b$ be a finite partition. Suppose $u,v:[a,b]\to\mathbb{R}$ are continuous and their restrictions to each $[t_{i-1},t_i]$ are absolutely continuous. Then $u,v$ are absolutely continuous on $[a,b]$, and $$ \int_a^b u(x)v'(x)\,dx = u(b)v(b)-u(a)v(a)-\int_a^b u'(x)v(x)\,dx. $$

Proof. First consider $u$. Given $\varepsilon>0$, on each of the finitely many pieces absolute continuity supplies a positive threshold for the total length of disjoint intervals lying in that piece, so that the sum of the corresponding changes in $u$ is less than $\varepsilon/N$. Take the minimum of these $N$ thresholds. Any interval in a disjoint family crossing a partition point can be split at each partition point it crosses. The resulting intervals lie in individual pieces; continuity at each partition point makes the changes over the split intervals add to the change over the original interval. If the total original length is below the minimum threshold, the total change is at most the sum of the bounds over the $N$ pieces, hence is less than or equal to $\varepsilon$; replacing $\varepsilon$ at the start by $\varepsilon/2$ gives the strict inequality required in the definition. Thus $u$ is absolutely continuous on $[a,b]$. The same argument applies to $v$. Now apply the finite-interval integration-by-parts formula on each $[t_{i-1},t_i]$ and sum. Each internal boundary term $u(t_i)v(t_i)$ appears once with a plus sign and once with a minus sign, so all internal terms cancel. The two terms left are $u(b)v(b)-u(a)v(a)$, proving the formula. $\square$

Worked Example: A Piecewise Linear Factor

On $[0,2]$ with Lebesgue measure, let $$ g(x)= \begin{cases} x,&0\le x\le1,\\ 2-x,&1\le x\le2. \end{cases} $$ This function is continuous, is absolutely continuous on each subinterval, and hence is absolutely continuous on $[0,2]$ by the proposition. Its derivative is $1$ on $(0,1)$ and $-1$ on $(1,2)$; the value at the single point $1$ does not affect the integral. Directly, $$ \int_0^2 e^{-x}g'(x)\,dx = \int_0^1 e^{-x}\,dx-\int_1^2 e^{-x}\,dx = (1-e^{-1})-(e^{-1}-e^{-2}) = 1-2e^{-1}+e^{-2}. $$ For a second calculation, use $u(x)=e^{-x}$, $v(x)=g(x)$. Both are absolutely continuous on $[0,2]$; $u'=-e^{-x}$, and $g(0)=g(2)=0$. Integration by parts gives $$ \int_0^2 e^{-x}g'(x)\,dx = \int_0^2 e^{-x}g(x)\,dx = \int_0^1 xe^{-x}\,dx+\int_1^2(2-x)e^{-x}\,dx. $$ An antiderivative of $xe^{-x}$ is $-(x+1)e^{-x}$, so the first integral is $1-2e^{-1}$. An antiderivative of $(2-x)e^{-x}$ is $(x-1)e^{-x}$, so the second is $e^{-2}-0=e^{-2}$. The result is $1-2e^{-1}+e^{-2}$, agreeing with the direct calculation.

Endpoint Checks and Common Pitfalls

On a finite interval, absolute continuity ensures the integration-by-parts formula applies, but the products at the endpoints still need to be evaluated. On an unbounded interval, the safe procedure is to work on $[0,R]$, check integrability of both integrands on $(0,\infty)$, and then check the limit of $u(R)v(R)$ as $R\to\infty$. The examples above do this through the exponential-power estimate and the lemma.

Common mistake: discarding the boundary term without checking it. In the cubic example, writing the formula directly on $[0,\infty)$ and simply deleting the term $-R^3e^{-2R}/2$ would leave an unjustified step. The correction is to retain it on $[0,R]$ and use the proved estimate to obtain $\lim_{R\to\infty}R^3e^{-2R}=0$. A different endpoint issue arises for $x^{-1/2}$: although $\int_0^1x^{-1/2}\,dx=2$, the function is unbounded at $0$, so it is not an absolutely continuous function on the closed interval $[0,1]$. The finite-interval formula cannot be applied to it there as if it were an AC factor. Integrability alone does not supply absolute continuity or justify an endpoint term.

At a join point, the derivative of a piecewise function may fail to exist, as it does for the tent-shaped $g$ at $x=1$. This causes no difficulty: the integration-by-parts formula uses derivatives defined almost everywhere, and a single point is a null set for Lebesgue measure. Continuity at the join is important because it makes the internal boundary terms cancel.

Check Your Understanding

For each question, state the measure and verify the relevant integrability or endpoint conditions before applying a formula.

Use the recurrence to compute $\int_0^\infty x^2e^{-4x}\,dx$, including the base integral and both recurrence steps.
For the piecewise function $g$ in the example, calculate $\int_0^2 e^{-x}g'(x)\,dx$ directly from its a.e. derivative.
Explain why $R^4e^{-3R}\to0$ and why $x^4e^{-3x}$ is integrable on $(0,\infty)$.
What continuity property at the partition point makes the internal boundary terms cancel in piecewise integration by parts?
Why does the finiteness of $\int_0^1x^{-1/2}\,dx$ not allow the AC integration-by-parts formula to be applied to $x^{-1/2}$ on $[0,1]$?

Worked Integration by Parts With Lebesgue Integrals

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