Term-by-Term Integration

Why Integration Requires Uniform Control

A finite partial sum of a power series is a polynomial, so its integral is found by integrating each term. For the infinite series, however, the sum is a limit of partial sums. To integrate term by term, we need to justify exchanging that limit with the integral. Uniform convergence supplies the needed control: it bounds the difference between the integral of a partial sum and the integral of its limit.

Inside the radius of convergence, the partial sums converge uniformly on every closed interval strictly inside the interval of convergence. This gives the term-by-term integration formula there. There is also an important boundary case: if the original power series converges at a boundary point, its partial sums converge uniformly along the segment from the center to that point. Thus integration term by term is valid at that endpoint as well.

Uniform Limits and Integration

Theorem (Integration of a Uniform Limit): Suppose $f_N:[u,v]\to\mathbb{R}$ are continuous and converge uniformly on $[u,v]$ to $f$. Then $f$ is continuous and $$ \lim_{N\to\infty}\int_u^v f_N(t)\,dt=\int_u^v f(t)\,dt. $$

Proof. Since each $f_N$ is continuous and the convergence is uniform, the Uniform Limits of Continuous Functions Are Continuous Theorem implies that $f$ is continuous. In particular, these functions are Riemann integrable on $[u,v]$. For every $N$, the standard integral estimate gives

\left|\int_u^v f_N(t)\,dt-\int_u^v f(t)\,dt\right| \leq \int_u^v |f_N(t)-f(t)|\,dt \leq (v-u)\sup_{t\in[u,v]}|f_N(t)-f(t)|.

The supremum on the right tends to zero by uniform convergence. Since $v-u$ is fixed, the difference between the integrals tends to zero, proving the result. If the endpoints are reversed, the same conclusion holds for oriented integrals because $\int_v^u=-\int_u^v$. $\square$

The estimate explains why pointwise convergence alone is not the hypothesis used here: it does not make the supremum of the errors tend to zero. Uniform convergence controls the error at every point of the interval at once.

Uniform Convergence Along a Boundary Segment

For a power series, convergence at a boundary point gives more than convergence at that single point. To see this, fix a boundary point $b$ with $|b-a|=R$, and suppose the series converges there. Put $d_n=c_n(b-a)^n$. The series $\sum_{n=0}^{\infty}d_n$ converges by assumption. Along the straight segment from $a$ to $b$, write $x=a+t(b-a)$ for $0\leq t\leq1$. The power series there becomes $\sum d_n t^n$.

Lemma (Uniform Convergence Along a Convergent Boundary Segment): If $\sum_{n=0}^{\infty}d_n$ converges, then $\sum_{n=0}^{\infty}d_n t^n$ converges uniformly for $0\leq t\leq1$.

Proof. Let $\varepsilon>0$. By the Cauchy Criterion for Series, there is an integer $N$ such that for all integers $q\geq p\geq N$, $$ \left|\sum_{n=p}^{q}d_n\right|<\varepsilon. $$ Fix such $p,q$, and set $A_k=\sum_{n=p}^{k}d_n$ for $p\leq k\leq q$. Summation by parts gives, for $0\leq t\leq1$,

\sum_{n=p}^{q}d_n t^n =A_qt^q+\sum_{n=p}^{q-1}A_n(t^n-t^{n+1}).

Every $A_n$ in this expression has absolute value less than $\varepsilon$. Also $t^n-t^{n+1}\geq0$, so

\left|\sum_{n=p}^{q}d_n t^n\right| \leq \varepsilon t^q+\varepsilon\sum_{n=p}^{q-1}(t^n-t^{n+1}) =\varepsilon t^p \leq\varepsilon.

The final bound is independent of $t$. Thus the partial sums satisfy the uniform Cauchy Criterion on $[0,1]$, and consequently converge uniformly there. $\square$

Applying the lemma with $d_n=c_n(b-a)^n$ proves uniform convergence of the power-series partial sums along the entire segment from $a$ to $b$. The argument also covers a boundary point to the left of the center: the parameterization $x=a+t(b-a)$ still has $0\leq t\leq1$, and the sign of $b-a$ is already included in $d_n$.

Term-by-Term Integration of a Power Series

Theorem (Term-by-Term Integration of a Power Series): Suppose $\sum_{n=0}^{\infty}c_n(x-a)^n$ has radius of convergence $R>0$, and let $f$ be its sum for $|x-a|<R$. For every such $x$, $$ \int_a^x f(t)\,dt =\sum_{n=0}^{\infty}\frac{c_n}{n+1}(x-a)^{n+1}. $$ If the original series also converges at a boundary point $b$, the same formula holds at $x=b$, with $f(b)$ defined by the convergent boundary series.

Proof. Let $S_N(t)=\sum_{n=0}^{N}c_n(t-a)^n$. First take $x$ strictly inside the radius, with $x\ne a$. The closed segment with endpoints $a$ and $x$ lies inside the interval of convergence. The Uniform Convergence Inside the Radius Theorem gives uniform convergence of $S_N$ to $f$ on that segment. The Integration of a Uniform Limit Theorem therefore gives

\int_a^x f(t)\,dt =\lim_{N\to\infty}\int_a^x S_N(t)\,dt =\lim_{N\to\infty}\sum_{n=0}^{N}\frac{c_n}{n+1}(x-a)^{n+1}.

The integral of each term follows from the power rule, including when $x<a$ because the integral is oriented. This proves the formula for interior $x$; at $x=a$, both sides are zero. If $b$ is a boundary point where the original series converges, the lemma just proved gives uniform convergence of $S_N$ along the segment from $a$ to $b$. Applying the Integration of a Uniform Limit Theorem on that segment gives the identical formula at $b$. $\square$

The integrated power series has the same radius of convergence as the original one. Its coefficients are shifted versions of $c_n/(n+1)$, and the factor $(n+1)^{-1}$ has $n$th root tending to $1$. The Cauchy–Hadamard Formula therefore gives the same coefficient growth rate, and hence the same radius. Integration changes the coefficients and raises each exponent by one, but does not enlarge the disk of convergence.

Worked Examples

Worked Example: Integrating the Geometric Series

For $|x|<1$, the geometric series has sum $\sum_{n=0}^{\infty}x^n=1/(1-x)$. Applying the theorem with center $a=0$ and coefficients $c_n=1$ gives

\int_0^x\frac{1}{1-t}\,dt =\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}, \qquad |x|<1.

Define $L(x)=\int_0^x\frac{1}{1-t}\,dt$ for $|x|<1$. Thus

L(x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots, \qquad |x|<1.

The integration theorem justifies this identity from uniform convergence on each segment from $0$ to an interior point $x$, rather than by informally integrating an infinite sum.

Worked Example: A Convergent Boundary Point

Consider $$ F(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}x^n. $$

The coefficient root limit is $1$, so the radius is $1$. At $x=1$, the series is the alternating harmonic series $\sum_{n=0}^{\infty}(-1)^n/(n+1)$, which converges by the Alternating Series Test. The boundary-segment lemma therefore gives uniform convergence of the partial sums on $[0,1]$, and term-by-term integration yields

\int_0^1 F(t)\,dt =\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)^2}.

Indeed, integrating the $n$th term gives $$ \int_0^1\frac{(-1)^n}{n+1}t^n\,dt =\frac{(-1)^n}{n+1}\cdot\frac{1}{n+1} =\frac{(-1)^n}{(n+1)^2}. $$ The original series diverges at $x=-1$, where it becomes $\sum 1/(n+1)$, so the boundary argument does not apply there. Boundary points must be checked individually.

Worked Example: Integrating a Series with a Finite Radius

Let $$ G(x)=\sum_{n=0}^{\infty}(n+1)\left(\frac{x+1}{2}\right)^n. $$

The ratio of consecutive coefficient magnitudes, when the series is written in powers of $x+1$, tends to $1/2$: $$ \frac{(n+2)/2^{n+1}}{(n+1)/2^n} =\frac{n+2}{2(n+1)} \longrightarrow\frac12. $$ Thus its radius about $a=-1$ is $2$. Set $z=(x+1)/2$. For $|z|<1$, differentiating the geometric-series identity, or multiplying the series by $1-z$, gives $\sum_{n=0}^{\infty}(n+1)z^n=1/(1-z)^2$. Term-by-term integration from $-1$ to $x$ gives

\int_{-1}^{x}G(t)\,dt =\sum_{n=0}^{\infty}\frac{n+1}{2^n}\frac{(x+1)^{n+1}}{n+1} =\sum_{n=0}^{\infty}\frac{(x+1)^{n+1}}{2^n}.

The last series is geometric in $z$, with first term $2z$, so its sum is $2z/(1-z)$. Direct integration confirms the result: since $dt=2\,dz$, $$ \int_{-1}^{x}\frac{1}{(1-(t+1)/2)^2}\,dt =2\int_0^z\frac{1}{(1-s)^2}\,ds =\frac{2z}{1-z}. $$ Both calculations are valid for $|x+1|<2$.

What the Boundary Result Does—and Does Not—Say

A useful distinction is that boundary convergence does justify term-by-term integration along the segment from the center to that boundary point. The boundary-segment lemma proves the required uniform convergence; it is not correct to assume that uniform convergence might fail on that segment when the endpoint series converges. Without convergence at the endpoint, however, that lemma provides no such uniform control, so pointwise convergence alone is not a justification for passing the integral through the limit.

This integration conclusion is specific to the segment and the integral in question. It does not assert uniform convergence on a whole interval extending past the radius, nor does it imply that the original series converges at the other boundary point. In the alternating example, convergence at $1$ supports the formula there, while divergence at $-1$ prevents the same argument at that endpoint.

Takeaway: On an interior segment, uniform convergence inside the radius permits integration term by term. If the power series converges at a boundary point, its partial sums also converge uniformly along the segment from the center to that point, so the integration formula holds there as well.

Check Your Understanding

Use the uniform-limit theorem and the power-series results in this tutorial to answer the following questions.

Which estimate bounds the difference between the integrals of a uniform limit and its approximating functions?
Why does convergence of $\sum d_n$ imply uniform convergence of $\sum d_n t^n$ for $0\leq t\leq1$?
What is the term-by-term integration formula for a power series centered at $a$?
What additional condition allows that formula to be used at a boundary point?
For the series $\sum_{n=0}^{\infty}(-1)^n x^n/(n+1)$, why does the argument apply at $x=1$ but not at $x=-1$?

What You'll Learn