Why Integrate a Power Series?
A power series can represent a function as an infinite sum of powers. If that representation is valid on an interval, integrating the series term by term can produce a new series for an antiderivative or a definite integral. This is especially useful when the antiderivative is not easy to express using elementary functions, or when a familiar function such as the arctangent or logarithm can be represented by a convenient series.
The basic calculation resembles the power rule for polynomials. The important additional question is where the infinite sum may be integrated this way. Inside the interval of convergence, the series converges uniformly on every closed subinterval, which justifies passing from integrals of finite sums to the integral of the infinite sum.
The Term-by-Term Integration Rule
Suppose a power series centered at \(a\) represents a function \(f\) for \(|x-a|<R\):
For any \(x\) strictly inside that interval, integrate from the center \(a\) to \(x\). The integral of each power is \(\frac{(x-a)^{n+1}}{n+1}\). Thus the resulting series has a new power of \(x-a\) and a denominator \(n+1\).
The variable \(t\) in the integral is a dummy variable: it indicates that the function is integrated with respect to its input. Using \(t\) also makes it clear that the resulting expression is a function of the endpoint \(x\).
Why Term-by-Term Integration Works
Write the partial sums of the original power series as \(S_N(t)=\sum_{n=0}^{N}c_n(t-a)^n\). For a finite sum, integration term by term is just linearity of the integral and the power rule. To justify the infinite version, we need the partial sums to approach \(f\) uniformly on the interval being integrated over.
Choose a closed interval \([u,v]\) strictly inside \((a-R,a+R)\). There is a number \(r<R\) such that \(|t-a|\le r\) for every \(t\in[u,v]\). Because the original power series converges absolutely at distance \(r\) from its center, \(\sum_{n=0}^{\infty}|c_n|r^n\) converges. For every \(t\in[u,v]\), the absolute value of the \(n\)th term is at most \(|c_n|r^n\). The tails of this convergent numerical series therefore bound the tails of the power series uniformly on \([u,v]\). Hence \(S_N\) converges uniformly to \(f\) there.
Uniform convergence means that the difference between \(S_N(t)\) and \(f(t)\) is small for all \(t\) in the interval at once. In particular, if the largest such difference is at most \(\varepsilon\), then:
As \(N\) increases, the uniform error \(\varepsilon\) tends to zero, so the integrals of the partial sums approach the integral of \(f\). Each partial sum is finite, and its integral is obtained term by term. This proves the term-by-term rule on closed intervals strictly inside the interval of convergence. Setting one endpoint equal to the center gives the displayed formula for \(\int_a^x f(t)\,dt\).
Worked Example: A Power Series for the Arctangent
Worked Example: Derive the Maclaurin Series for \(\arctan x\)
For \(|t|<1\), the geometric series with input \(-t^2\) gives:
Since the derivative of \(\arctan t\) is \(\frac{1}{1+t^2}\), integrate from \(0\) to \(x\). For \(|x|<1\), the interval between \(0\) and \(x\) lies inside the convergence interval, so term-by-term integration applies:
Because \(\arctan 0=0\), the left side is \(\arctan x\). The first terms are \(x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}\). To check the result, differentiate the series inside \(|x|<1\): its derivative is \(1-x^2+x^4-x^6+\cdots=\frac{1}{1+x^2}\), the derivative of \(\arctan x\). At \(x=0\), both the series and \(\arctan x\) equal \(0\), so the constant of integration also agrees.
Worked Example: A Power Series for \(\ln(1+x)\)
Worked Example: Integrate the Geometric Series to Find \(\ln(1+x)\)
For \(|t|<1\), substituting \(-t\) into the geometric series gives:
The derivative of \(\ln(1+t)\) is \(\frac{1}{1+t}\) when \(1+t>0\). In particular, for \(-1<t<1\), integrate from \(0\) to \(x\):
Since \(\ln(1)=0\), this is the series for \(\ln(1+x)\). Its first terms are \(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}\). As a check, differentiating the series gives \(1-x+x^2-x^3+\cdots=\frac{1}{1+x}\) for \(|x|<1\), and the series has value \(0\) at \(x=0\), just like \(\ln(1+x)\).
Worked Example: Represent a Definite Integral Exactly
Worked Example: Express \(\ln 2\) as an Infinite Series
We can use the geometric series for \(\frac{1}{1-t}\) to represent the integral that equals \(\ln 2\). For \(0\le t\le\frac12\), the geometric series converges, and this whole interval lies strictly inside its interval of convergence:
Integrate from \(0\) to \(\frac12\), integrating each term:
Evaluating the integral directly gives \([-\ln(1-t)]_0^{1/2}=-\ln(\frac12)+\ln(1)=\ln 2\). Therefore:
To check the series, integrate the finite partial sum \(1+t+t^2+\cdots+t^N\). Its integral from \(0\) to \(\frac12\) is \(\sum_{n=0}^{N}\frac{(1/2)^{n+1}}{n+1}\). As \(N\) increases, the partial sums converge uniformly on this interval to \(\frac{1}{1-t}\), so these integrals approach \(\int_0^{1/2}\frac{1}{1-t}\,dt=\ln 2\), as claimed.
Constants, Definite Integrals, and a Common Mistake
Integration of a series produces a family of antiderivatives, so the constant \(C\) must be included when a general antiderivative is requested. Integrating from the center instead fixes the constant automatically: \(\int_a^a f(t)\,dt=0\). For a definite integral, integrate each term over the stated bounds and evaluate the resulting powers at both endpoints.
A frequent error is to increase the exponent but leave the old exponent in the denominator. For instance, the incorrect calculation \(\int x^3\,dx=\frac{x^4}{3}\) fails because differentiating the proposed answer gives \(\frac{4x^3}{3}\), not \(x^3\). The correct calculation is:
The same indexing rule applies to a power series: \(c_n(x-a)^n\) integrates to \(\frac{c_n}{n+1}(x-a)^{n+1}\), not to a term with denominator \(n\). Term-by-term integration is guaranteed on intervals strictly inside the original interval of convergence. At an endpoint, convergence must be checked separately; the interior rule alone does not settle what happens there.
Check Your Understanding
Use the term-by-term integration rule and state the interval on which your reasoning applies.
- Integrate \(\sum_{n=0}^{\infty}x^n\) term by term from \(0\) to \(x\), for \(|x|<1\).
- Starting from \(\frac{1}{1+t^2}=\sum_{n=0}^{\infty}(-1)^nt^{2n}\), write the first four nonzero terms of the integrated series.
- What constant must be added to an antiderivative series if it is required to equal \(5\) at its center \(x=a\)?
- Explain why uniform convergence on a closed interval inside the interval of convergence justifies integrating the series there.
- Identify the error in the claim \(\int c_n(x-a)^n\,dx=\frac{c_n}{n}(x-a)^{n+1}\), and write the corrected term.
