Integration by Parts for Absolutely Continuous Functions

Why Integration by Parts Extends Beyond Smooth Functions

For continuously differentiable functions, integration by parts follows by integrating the product rule. Absolute continuity gives a broader setting: the functions need not have continuous derivatives, or even derivatives at every point. The key is that an absolutely continuous function is recovered from its integrable derivative by an integral. We will show that the product remains absolutely continuous, apply the product rule wherever both derivatives exist, and then use that integral representation.

Throughout, $a<b$ are finite real numbers, and the measure space is $([a,b],\mathcal{L}_{[a,b]},m)$, where $\mathcal{L}_{[a,b]}$ is the restriction of the Lebesgue sigma-algebra and $m$ is Lebesgue measure. This is a finite, complete measure space, with $m([a,b])=b-a$. All functions in the theorem are real-valued. An equality involving derivatives is asserted $m$-almost everywhere; the final integration-by-parts identity is an equality of real numbers.

Definition: A real-valued function $f$ on $[a,b]$ is absolutely continuous if, for every $\varepsilon>0$, there is a $\delta>0$ such that for every finite collection of pairwise disjoint intervals $(u_j,v_j)\subseteq[a,b]$, the condition $\sum_j(v_j-u_j)<\delta$ implies $\sum_j|f(v_j)-f(u_j)|<\varepsilon$.

We use the integral representation theorem for absolutely continuous functions: if $f$ is real-valued and absolutely continuous on $[a,b]$, then $f'$ exists $m$-almost everywhere, is Lebesgue integrable, and $f(x)=f(a)+\int_a^x f'(t)\,dt$ for every $x\in[a,b]$. Conversely, an indefinite Lebesgue integral of an integrable function, plus a constant, is absolutely continuous. This is the standard integral representation theorem established in the earlier material on absolute continuity. In particular, an absolutely continuous function is continuous and hence bounded on $[a,b]$.

The Formula and Its Proof

Theorem (Integration by Parts): On the finite Lebesgue measure space $([a,b],\mathcal{L}_{[a,b]},m)$, let $f,g:[a,b]\to\mathbb{R}$ be absolutely continuous. Then $f'g$ and $fg'$ are Lebesgue integrable and $$ \int_a^b f(x)g'(x)\,dx =f(b)g(b)-f(a)g(a)-\int_a^b f'(x)g(x)\,dx. $$ The derivatives in the integrands can be assigned arbitrary values on the null sets where they do not exist; the integrals do not depend on those assignments.

Proof. First we show that $h=fg$ is absolutely continuous. Since $f$ and $g$ are continuous on the compact interval $[a,b]$, the numbers $M_f=\sup_{[a,b]}|f|$ and $M_g=\sup_{[a,b]}|g|$ are finite. For any $u,v\in[a,b]$, $$ |f(v)g(v)-f(u)g(u)| \le M_f|g(v)-g(u)|+M_g|f(v)-f(u)|. $$ Indeed, write the difference as $f(v)(g(v)-g(u))+g(u)(f(v)-f(u))$ and apply the triangle inequality.

Fix $\varepsilon>0$ and put $\eta=\varepsilon/(2(M_f+M_g+1))$. Absolute continuity of $f$ and $g$ supplies positive numbers $\delta_f,\delta_g$ such that either function has total change less than $\eta$ across any finite disjoint collection of intervals whose total length is less than its corresponding $\delta$. Set $\delta=\min(\delta_f,\delta_g)$. For disjoint intervals $(u_j,v_j)$ with total length less than $\delta$, summing the displayed inequality gives $$ \sum_j|h(v_j)-h(u_j)| \le M_f\sum_j|g(v_j)-g(u_j)|+M_g\sum_j|f(v_j)-f(u_j)| <(M_f+M_g)\eta<\varepsilon. $$ Thus $h$ is absolutely continuous.

By the integral representation theorem, $f'$ and $g'$ exist almost everywhere. Outside the union of their two exceptional null sets, both functions are differentiable, so the ordinary product rule gives $h'=f'g+fg'$. The union is null because it is a finite union of null sets. Moreover, $f$ and $g$ are bounded and $f',g'$ are integrable, so $$ |f'g+fg'|\le M_g|f'|+M_f|g'| $$ almost everywhere, and the right-hand side is integrable. Apply the integral representation theorem to the absolutely continuous function $h$. It gives $h(b)-h(a)=\int_a^b h'(x)\,dx$. Substituting the product-rule identity, which holds almost everywhere, yields $$ f(b)g(b)-f(a)g(a) =\int_a^b f'(x)g(x)\,dx+\int_a^b f(x)g'(x)\,dx. $$ Rearranging proves the formula and also shows both integrals are finite. $\square$

If $a=b$, both integrals over the degenerate interval are zero and $f(b)g(b)-f(a)g(a)=0$, so the identity holds in that case as well. The proof for $a<b$ explains why no continuity of the derivatives is needed: the product rule is used almost everywhere, and the integral representation theorem converts that derivative identity into an endpoint identity.

Worked Applications

Worked Example: A Smooth Function Paired With an Exponential

On $[0,1]$ with Lebesgue measure, take $f(x)=x^2$ and $g(x)=e^x$. Both are continuously differentiable, hence absolutely continuous. Their derivatives are $f'(x)=2x$ and $g'(x)=e^x$. Apply integration by parts to compute $\int_0^1x^2e^x\,dx$:

\begin{aligned} \int_0^1x^2e^x\,dx &=[x^2e^x]_0^1-\int_0^1 2xe^x\,dx\\ &=e-\left[2(x-1)e^x\right]_0^1\\ &=e-\bigl(0-(-2)\bigr)\\ &=e-2. \end{aligned}

Here $[x^2e^x]_0^1=1^2e^1-0^2e^0=e$, and $[2(x-1)e^x]_0^1=2(1-1)e^1-2(0-1)e^0=0-(-2)=2$. Check: differentiating $e^x(x^2-2x+2)$ gives $e^x(x^2-2x+2)+e^x(2x-2)=x^2e^x$. Its endpoint difference is $e(1-2+2)-1(0-0+2)=e-2$, as required.

Worked Example: A Nondifferentiable Point

On $[-1,1]$ with Lebesgue measure, let $f(x)=|x|$ and $g(x)=x$. The function $f$ is Lipschitz, and thus absolutely continuous; $g$ is also absolutely continuous. The derivative $f'$ is $-1$ for $x<0$ and $1$ for $x>0$, and is not defined at $0$. This single point is a null set, so it does not affect the integrals. Since $g'=1$,

\int_{-1}^{1}|x|\,dx =1-(-1)-\int_{-1}^{1}f'(x)x\,dx =2-\int_{-1}^{1}|x|\,dx.

In the last equality, $f'(x)x=|x|$ for every $x\ne0$. Directly, $\int_{-1}^1|x|\,dx=2\int_0^1x\,dx=2(1/2)=1$. The formula therefore gives $2-1=1$. Check: the endpoint products are $f(1)g(1)=1$ and $f(-1)g(-1)=1\cdot(-1)=-1$, whose difference is $2$; subtracting the integral $1$ gives $1$, matching the direct computation.

Worked Example: An Integrable Derivative That Is Unbounded

On $[0,1]$ with Lebesgue measure, define $f(x)=2\sqrt{x}$ and $g(x)=x$. For $x>0$, $f'(x)=1/\sqrt{x}$, which is unbounded near $0$, but it is integrable. Indeed, $f(x)=\int_0^x t^{-1/2}\,dt$, since that integral equals $2\sqrt{x}$; the converse part of the integral representation theorem therefore gives that $f$ is absolutely continuous. The function $g$ is absolutely continuous and $g'=1$. The formula yields

\begin{aligned} \int_0^1 2\sqrt{x}\,dx &=[2x^{3/2}]_0^1-\int_0^1\frac{x}{\sqrt{x}}\,dx\\ &=2-\int_0^1\sqrt{x}\,dx\\ &=2-\left[\frac{2}{3}x^{3/2}\right]_0^1\\ &=2-\frac23=\frac43. \end{aligned}

The quotient $x/\sqrt{x}=\sqrt{x}$ holds for $x>0$, which suffices for equality almost everywhere. Check: the left side is $[\,\frac43x^{3/2}\,]_0^1=4/3$, independently confirming the result. This example shows that the theorem does not require bounded derivatives; integrability of the derivatives is the relevant property supplied by absolute continuity.

A Sign Pitfall and the Role of the Hypotheses

The identity can be remembered in the symmetric form $$ \int_a^b \bigl(f'(x)g(x)+f(x)g'(x)\bigr)\,dx =f(b)g(b)-f(a)g(a). $$ To isolate $\int f g'$, the term $\int f'g$ must be subtracted, not added. For a concrete check on $[0,1]$ with Lebesgue measure, set $f(x)=g(x)=x$. The correct calculation is $$ \int_0^1 x\cdot1\,dx=[x^2]_0^1-\int_0^1 1\cdot x\,dx =1-\frac12=\frac12. $$ The incorrect sign would give $1+\frac12=\frac32$, which contradicts the direct integral $\int_0^1x\,dx=1/2$.

The hypotheses matter for the proof. Absolute continuity ensures that the functions have integrable derivatives and are recovered from them by the integral representation theorem. Differentiability almost everywhere alone does not supply that representation. For example, the Cantor function is continuous and has derivative zero almost everywhere but is not constant; it is not absolutely continuous. Thus one cannot infer an endpoint difference solely by integrating an almost-everywhere derivative identity unless an appropriate integral representation is available.

For absolutely continuous functions, integration by parts is therefore not merely a formal manipulation of derivatives. The product is absolutely continuous, its derivative obeys the product rule almost everywhere, and its change between endpoints is exactly the Lebesgue integral of that derivative. These facts together justify the formula even when derivatives are discontinuous, unbounded, or undefined on a null set.

Check Your Understanding

For each question, specify the interval and verify that the functions satisfy the hypotheses before applying the formula.

On $[0,1]$, use integration by parts to compute $\int_0^1 x\cos x\,dx$.
On $[0,2]$, take $f(x)=x$ and $g(x)=x^2$. Compute both integrals on the right side of the symmetric identity and check the endpoint difference.
Explain why an undefined derivative at a single point does not affect the integration-by-parts formula.
In the product proof, identify where boundedness of $f$ and $g$ is used and where absolute continuity is used.
For $f(x)=g(x)=x$ on $[0,1]$, write the incorrect result obtained by adding rather than subtracting $\int_0^1 f'g\,dx$, and compare it with the direct integral.

Integration by Parts for Absolutely Continuous Functions

What You'll Learn