Why Integrate a Power Series on an Interval?
A power series can describe a function throughout the interval where it converges. If we want the definite integral of that function over a smaller closed interval, it is natural to try integrating each term and then adding the results. The important question is whether this procedure really gives the integral of the sum.
The key condition is that the whole interval lie strictly inside the series's radius of convergence. There, the power series converges uniformly, which lets us control the error between the function and its partial sums across the entire interval. That uniform control justifies passing from integrals of partial sums to the integral of the function.
The Setting and the Formula
Consider a power series centered at \(c\), with real coefficients \(a_n\) and radius of convergence \(R>0\):
Let \(a\le b\), and suppose the closed interval \([a,b]\) lies strictly inside the interval of convergence. Equivalently, its most distant endpoint from \(c\) is still less than \(R\):
The maximum occurs at an endpoint because the distance \(|x-c|\) cannot exceed the larger endpoint distance for any \(x\in[a,b]\). Under this condition, the definite-integral formula is:
Each term on the right is obtained by integrating \(a_n(x-c)^n\) from \(a\) to \(b\). The theorem says that the infinite sum of those term integrals equals the integral of the function represented by the power series.
Why Uniform Convergence Justifies the Formula
For \(N\ge 0\), define the \(N\)th partial sum by \(P_N(x)=\sum_{n=0}^{N}a_n(x-c)^n\). Because \(\rho<R\), the series \(\sum_{n=0}^{\infty}|a_n|\rho^n\) converges. For every \(x\in[a,b]\), \(|x-c|\le\rho\), so
The Weierstrass M-test therefore gives uniform convergence of the power series on \([a,b]\). In particular, its partial sums approach \(f\) uniformly. Here is how that fact gives the integral identity.
Uniform convergence means that for every \(\varepsilon>0\), some \(N_0\) makes \(|f(x)-P_N(x)|<\varepsilon\) for every \(x\in[a,b]\) whenever \(N\ge N_0\).
For \(N\ge N_0\), the integral inequality gives \(\left|\int_a^b f(x)\,dx-\int_a^b P_N(x)\,dx\right|\le (b-a)\varepsilon\).
Linearity of integration gives \(\int_a^b P_N(x)\,dx=\sum_{n=0}^{N}a_n\frac{(b-c)^{n+1}-(a-c)^{n+1}}{n+1}\).
As \(N\) tends to infinity, the integral error tends to zero. Thus the integrals of the partial sums converge to \(\int_a^b f(x)\,dx\), proving the formula.
This argument proves the needed result directly: it does not merely rely on formally writing an integral sign in front of an infinite sum. The condition \(\rho<R\) is what supplies uniform convergence on the full interval.
Worked Example: Integrating the Exponential Series
Worked Example: Integrating the Exponential Series
The exponential function has the power series \(e^x=\sum_{n=0}^{\infty}x^n/n!\), which has infinite radius of convergence. Evaluate \(\int_0^1 e^x\,dx\) by integrating its series term by term.
The interval \([0,1]\) is strictly inside the radius \(R=\infty\). The theorem applies, with \(c=0\), and gives
Substituting the first four values of \(n\) gives \(1/(1\cdot0!)=1\), \(1/(2\cdot1!)=1/2\), \(1/(3\cdot2!)=1/6\), and \(1/(4\cdot3!)=1/24\). Thus the series begins
For a check, the antiderivative of \(e^x\) is \(e^x\), so direct evaluation gives \(e^1-e^0=e-1\). The integrated power series also sums to \(e-1\), since \(\sum_{n=0}^{\infty}1/((n+1)n!)=\sum_{m=1}^{\infty}1/m!=e-1\), where \(m=n+1\). The two evaluations agree.
Worked Example: Integrating the Sine Series
Worked Example: Integrating the Sine Series
Use the power series \(\sin x=\sum_{k=0}^{\infty}(-1)^k x^{2k+1}/(2k+1)!\) to evaluate \(\int_0^1\sin x\,dx\).
This series has infinite radius of convergence, so term-by-term integration is valid on \([0,1]\). Integrating its \(k\)th term gives
Therefore
For a check, an antiderivative of \(\sin x\) is \(-\cos x\). Evaluating at the endpoints gives \(-\cos(1)-(-\cos(0))=1-\cos(1)\). The cosine series gives \(1-\cos(1)=1-(1-1/2!+1/4!-1/6!+\cdots)\), which is exactly the displayed integrated series.
Worked Example: A Series with a Finite Radius
Worked Example: A Series with a Finite Radius
Consider \(f(x)=\sum_{n=0}^{\infty}(x-1)^n/(n+1)^2\), whose radius of convergence is \(1\). Use term-by-term integration to estimate \(\int_1^{3/2}f(x)\,dx\) to three decimal places.
The interval has center \(c=1\), and its most distant point from the center is \(1/2\). Since \(1/2<1\), the closed interval is strictly inside the radius of convergence, so the theorem applies. For each term,
Consequently, the integral equals \(\sum_{n=0}^{\infty}(1/2)^{n+1}/(n+1)^3\). The first four terms, for \(n=0,1,2,3\), are \(1/2\), \(1/32\), \(1/216\), and \(1/1024\). Their sum is
To control the omitted terms, for \(n\ge4\) we have \(1/(n+1)^3\le1/125\). Hence the remaining tail is bounded by
The integral therefore lies between \(0.5368561921\ldots\) and \(0.5373561921\ldots\), so it rounds to \(0.537\) to three decimal places. As a check on the integrated series, its power series in \(x\) is \(\sum_{n=0}^{\infty}(x-1)^{n+1}/((n+1)^3)\). Differentiating term by term for \(|x-1|<1\) gives \(\sum_{n=0}^{\infty}(x-1)^n/(n+1)^2=f(x)\), and the integrated series is zero at \(x=1\), as required for the integral from \(1\).
A Common Mistake: Integrating the Coefficients but Not the Powers
A frequent error is to integrate each term but forget that the exponent increases by one and the coefficient is divided by the new exponent. For instance, the incorrect calculation
does not produce an antiderivative: differentiating its right-hand side gives \(\sum_{n=0}^{\infty}(n+1)x^n/n!\), not \(\sum_{n=0}^{\infty}x^n/n!\). The correct term-by-term antiderivative is
because differentiating each term gives \(x^n/n!\). For a definite integral, the corresponding term is the endpoint difference divided by \(n+1\), as in the theorem.
A second issue is applying the theorem when an endpoint lies on the radius boundary. At such a point, the hypothesis \(\rho<R\) fails, so the uniform-convergence argument given here does not apply. This does not by itself prove that term-by-term integration is false there; it means the interior-interval theorem cannot justify it. Boundary cases require separate conditions and reasoning.
Why the Interior Condition Is Useful
The interior condition gives more than a formal way to manipulate a series. It provides a reliable route from a function represented by a power series to a definite integral, and it lets us calculate that integral from coefficients. In examples, the resulting series can also be approximated with explicit error bounds, as in the finite-radius example.
When applying the result, check three points:
- Identify the center \(c\), coefficients \(a_n\), and radius \(R\).
- Confirm that both endpoints satisfy \(|a-c|<R\) and \(|b-c|<R\).
- Integrate each term with the factor \(1/(n+1)\), then evaluate at both endpoints.
The strict endpoint check is essential: a series may converge at some boundary points without converging uniformly on a closed interval that reaches the boundary. On a closed interval strictly inside the radius, uniform convergence supplies the needed justification.
Check Your Understanding
Use the theorem and its hypotheses to answer these questions.
- For a power series centered at \(c\) with radius \(R\), what inequality must \(\max\{|a-c|,|b-c|\}\) satisfy before the interior-interval theorem applies?
- Why does uniform convergence on \([a,b]\) allow the integrals of the partial sums to approach the integral of the represented function?
- Write the term-by-term formula for \(\int_a^b\sum_{n=0}^{\infty}a_n(x-c)^n\,dx\).
- What is the \(n=2\) integrated term in \(\int_0^1\sum_{n=0}^{\infty}x^n/n!\,dx\)?
- If one endpoint lies exactly at distance \(R\) from the center, does the interior-interval theorem apply? Explain what the failed condition does—and does not—tell you.