Term-by-Term Integration of Function Series

From Function Series to Series of Integrals

A function series has partial sums $S_N(x)=\sum_{n=1}^{N}f_n(x)$. If these partial sums converge uniformly on an interval, their limit is controlled across the entire interval, rather than only at each fixed input. This control is what allows integration to pass from the partial sums to their limit. Since each partial sum is a finite sum, its integral is already the sum of the integrals of its terms. The key question is whether that identity persists as $N$ tends to infinity.

The result below concerns Riemann integration on a closed, bounded interval. Uniform convergence is a sufficient hypothesis for term-by-term integration; absolute uniform convergence is not required. We will also see why replacing uniform convergence with pointwise convergence can lead to a wrong answer.

Definition: Let $a<b$, and suppose each $f_n:[a,b]\to\mathbb{R}$ is Riemann integrable. If the series $\sum_{n=1}^{\infty}f_n(x)$ converges uniformly on $[a,b]$ to $S(x)$, its term-by-term integral is the numerical series $$ \sum_{n=1}^{\infty}\int_a^b f_n(x)\,dx. $$ Term-by-term integration means that this series converges and that its sum equals $\int_a^b S(x)\,dx$.

The definition describes a conclusion, not a guarantee: a term-by-term integral is justified only when an appropriate theorem applies. In this setting, uniform convergence supplies the needed guarantee. The role of the interval matters too: the statements below concern Riemann integrals on $[a,b]$, not improper integrals on unbounded intervals.

The Term-by-Term Integration Theorem

Theorem (Term-by-Term Integration of a Uniformly Convergent Series): Let $a<b$. Suppose every $f_n:[a,b]\to\mathbb{R}$ is Riemann integrable and $\sum_{n=1}^{\infty}f_n(x)$ converges uniformly on $[a,b]$ to $S(x)$. Then $S$ is Riemann integrable, the numerical series of integrals converges, and $$ \int_a^b S(x)\,dx =\sum_{n=1}^{\infty}\int_a^b f_n(x)\,dx. $$

Proof. Write $S_N=\sum_{n=1}^{N}f_n$. Every $S_N$ is Riemann integrable because it is a finite sum of Riemann integrable functions. By hypothesis, $S_N$ converges uniformly to $S$. The earlier result Uniform Limits of Riemann Integrable Functions Are Riemann Integrable therefore gives that $S$ is Riemann integrable. By Convergence of Integrals under Uniform Convergence, $$ \lim_{N\to\infty}\int_a^b S_N(x)\,dx=\int_a^b S(x)\,dx. $$ For every finite $N$, linearity of the Riemann integral gives $$ \int_a^b S_N(x)\,dx =\int_a^b\left(\sum_{n=1}^{N}f_n(x)\right)\,dx =\sum_{n=1}^{N}\int_a^b f_n(x)\,dx. $$ Thus the partial sums of the numerical series of integrals converge to $\int_a^b S(x)\,dx$. This proves both the convergence of that series and the asserted equality. $\square$

The proof separates the argument into two parts. Uniform convergence allows the integral of the limit to be obtained as the limit of the integrals of the partial sums. Linearity then identifies each of those finite integrals with a finite sum of term integrals. No infinite sum is moved through an integral without first taking this limit.

Worked Example: Integrating a Geometric Series

Fix $r$ with $0<r<1$. On $[0,r]$, let $f_n(x)=x^n$ for $n\geq0$. Each term is continuous, and $$ |f_n(x)|=x^n\leq r^n\qquad(0\leq x\leq r). $$ Since $\sum_{n=0}^{\infty}r^n$ converges, the Weierstrass M-Test gives uniform convergence of $\sum_{n=0}^{\infty}x^n$ on $[0,r]$. Its sum is $S(x)=1/(1-x)$: the finite geometric-sum identity gives $$ \sum_{n=0}^{N}x^n=\frac{1-x^{N+1}}{1-x}, $$ and $x^{N+1}\leq r^{N+1}\to0$ uniformly on this interval.

The term-by-term integration theorem now yields $$ \int_0^r\frac{1}{1-x}\,dx =\sum_{n=0}^{\infty}\int_0^r x^n\,dx =\sum_{n=0}^{\infty}\frac{r^{n+1}}{n+1}. $$ The left side is $-\ln(1-r)$. Equivalently, this agrees with the earlier Logarithm Series after substituting $-r$ for its variable. The integration theorem justifies the equality between the integral and the infinite series; the uniform bound is the essential hypothesis behind that step.

A Uniform Bound for the Integration Error

The theorem gives an equality of limits, but its proof also suggests a practical way to estimate how accurately a finite number of integrated terms approximates the full integral. The error is exactly the integral of the remainder $S-S_N$. If that remainder is uniformly small, its integral is small as well.

Theorem (Integrated Remainder Estimate): Under the hypotheses of the Term-by-Term Integration Theorem, for every $N\geq1$, $$ \left|\int_a^b S(x)\,dx-\sum_{n=1}^{N}\int_a^b f_n(x)\,dx\right| \leq (b-a)\sup_{x\in[a,b]}|S(x)-S_N(x)|. $$

Proof. Finite linearity gives $\sum_{n=1}^{N}\int_a^b f_n=\int_a^b S_N$, so the expression inside the absolute value is $\int_a^b(S-S_N)$. For any Riemann integrable function $h$, monotonicity of the integral and the pointwise inequalities $-|h|\leq h\leq |h|$ imply $$ \left|\int_a^b h(x)\,dx\right|\leq\int_a^b|h(x)|\,dx. $$ If $M=\sup_{x\in[a,b]}|h(x)|$, then $|h(x)|\leq M$, and monotonicity also gives $\int_a^b|h(x)|\,dx\leq(b-a)M$. Apply these inequalities to $h=S-S_N$, which is Riemann integrable, to obtain the stated estimate. $\square$

Because $S_N$ converges uniformly to $S$, the supremum on the right tends to zero. The estimate makes the convergence of the integrated partial sums quantitative: if the uniform remainder is at most $\delta$, the error in the integral is at most $(b-a)\delta$. A long interval magnifies the possible error by its length, which is why the length factor should not be omitted.

Worked Example: A Uniform Series That Is Not Absolutely Uniform

On $[0,1]$, define $$ f_n(x)=(-1)^{n+1}\frac{x^{n-1}}{n},\qquad n\geq1. $$ For each fixed $x\in[0,1]$, the magnitudes $a_n(x)=x^{n-1}/n$ are nonnegative and nonincreasing. Indeed, $$ \frac{a_{n+1}(x)}{a_n(x)}=\frac{xn}{n+1}\leq\frac{n}{n+1}<1 $$ when $a_n(x)>0$; at $x=0$, all terms after the first are zero. The alternating-series remainder estimate therefore gives, for every $x\in[0,1]$, $$ \left|\sum_{n=N+1}^{\infty}f_n(x)\right| \leq a_{N+1}(x) =\frac{x^N}{N+1} \leq\frac{1}{N+1}. $$ This bound tends to zero independently of $x$, so the series converges uniformly.

It is not absolutely uniformly convergent: at $x=1$, its series of absolute values is $\sum_{n=1}^{\infty}1/n$, which diverges. Nevertheless, term-by-term integration applies because uniform convergence is enough. Since $$ \int_0^1 f_n(x)\,dx =\frac{(-1)^{n+1}}{n}\int_0^1x^{n-1}\,dx =\frac{(-1)^{n+1}}{n^2}, $$ we obtain $$ \int_0^1 S(x)\,dx =\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}. $$ This example illustrates that the integration theorem does not require absolute uniform convergence. The signed series may converge uniformly even when the absolute series does not.

Why Pointwise Convergence Does Not Suffice

Pointwise convergence controls the partial sums separately at each input, but it does not ensure that their integrals approach the integral of their pointwise limit. The next example constructs continuous partial sums whose pointwise limit is zero, even though every partial sum has integral one.

Worked Example: Pointwise Convergence with the Wrong Integral Limit

For each $n\geq1$, set $a_n=1/(n+1)$, $b_n=1/n$, and $L_n=b_n-a_n=1/(n(n+1))$. Define a triangular function $g_n$ to be zero outside $[a_n,b_n]$, and on that interval define $$ g_n(x)=\frac{4}{L_n^2}\min(x-a_n,b_n-x). $$ This function is continuous, is zero at both endpoints, and has peak height $2/L_n$ at the midpoint. Its graph is a triangle with base $L_n$ and height $2/L_n$, so $$ \int_0^1g_n(x)\,dx=\frac{L_n(2/L_n)}{2}=1. $$ The intervals $[a_n,b_n]$ have disjoint interiors. At any fixed $x$, at most one of the $g_n(x)$ is nonzero, so $g_n(x)\to0$ pointwise.

Now put $g_0=0$ and define $f_n=g_n-g_{n-1}$ for $n\geq1$. Each $f_n$ is continuous. The partial sums telescope: $$ \sum_{k=1}^{N}f_k(x)=g_N(x)-g_0(x)=g_N(x). $$ Consequently, the series $\sum_{n=1}^{\infty}f_n(x)$ converges pointwise to the zero function, whose integral is zero. But each partial sum has integral one: $$ \int_0^1\sum_{k=1}^{N}f_k(x)\,dx =\int_0^1g_N(x)\,dx =1. $$ Equivalently, $\int_0^1f_1=1$, while $\int_0^1f_n=1-1=0$ for $n\geq2$. Thus the series of term integrals sums to one, not to the integral of the pointwise limit. There is no contradiction with the theorem: the partial sums do not converge uniformly to zero.

A common pitfall is to integrate each term and assume that the resulting series must give the integral of the pointwise sum. The triangular functions show why this is unsafe: their supports shrink, but their heights grow so that every partial sum retains the same area. Uniform convergence rules out this behavior by controlling the remainder everywhere on the interval.

Check integrability.
Verify that each term is Riemann integrable on the closed interval.

Establish uniform convergence.
Use a uniform tail estimate, the Weierstrass M-Test, or another valid uniform convergence argument.

Integrate the partial sums.
Apply finite linearity first, then use the theorem to pass to the limit.

Estimate the remainder if needed.
Multiply a uniform bound for $S-S_N$ by the interval length to bound the integral error.

Takeaway: On a closed, bounded interval, a uniformly convergent series of Riemann integrable functions may be integrated term by term. Uniform convergence also bounds the integration error, while pointwise convergence alone does not justify interchanging the series and the integral.

Check Your Understanding

Use the hypotheses and estimates in this tutorial to answer the following questions.

What two conditions on the terms and their series are needed for the term-by-term integration theorem?
Why is it legitimate to integrate each finite partial sum before passing to the limit?
How does a uniform bound for the remainder control the error in the integral?
Why does the alternating power series example qualify for term-by-term integration even though it is not absolutely uniformly convergent?
In the triangular-function example, what prevents the pointwise limit from determining the limit of the integrals?

Term-by-Term Integration of Function Series

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