How to Solve Integrals Using Series

A power series can turn a difficult integral into a sequence of elementary integrals. The basic idea is to replace the integrand with a sum of powers, integrate those powers, and then evaluate or approximate the resulting series.

The key caution is convergence: a series can represent a function only on part of the real line, and integrating a series term by term requires justification. The method is especially useful when the integrand has a familiar series expansion but no convenient elementary antiderivative.

The basic method

Suppose the integrand has a power-series expansion

$$f(x)=\sum_{n=0}^{\infty} c_n(x-a)^n.$$

On an interval where you can integrate term by term, an antiderivative is

$$\int f(x)\,dx = C+\sum_{n=0}^{\infty}\frac{c_n}{n+1}(x-a)^{n+1}.$$

For a definite integral, integrate each term between the limits:

$$\int_\alpha^\beta f(x)\,dx =\sum_{n=0}^{\infty}c_n\int_\alpha^\beta(x-a)^n\,dx.$$

A reliable workflow is:

1. Identify a known series that matches the integrand or a part of it.
2. Find where that series converges and where it equals the original function.
3. Check whether the full integration interval lies within that region.
4. Integrate the terms and evaluate the resulting sum, or use partial sums for an approximation.

Useful expansions include the geometric series $1/(1-u)=\sum_{n=0}^\infty u^n$ for $|u|<1$, and the Taylor series for $e^x$, $\sin x$, and $\cos x$, which converge for every real $x$.

Worked example: integrate a rational function

Evaluate

$$I=\int_0^1\frac{x}{1+x^2}\,dx.$$

For $|x|<1$, the geometric series gives

$$\frac{x}{1+x^2}=x\frac{1}{1-(-x^2)} =\sum_{n=0}^{\infty}(-1)^n x^{2n+1}.$$

Integrating the terms from $0$ to $1$ suggests

$$I=\sum_{n=0}^{\infty}(-1)^n\int_0^1 x^{2n+1}\,dx =\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+2}.$$

The series for the integrand is not convergent absolutely at $x=1$, so it is worth checking the endpoint rather than assuming term-by-term integration automatically applies. Let

$$S_N(x)=\sum_{n=0}^{N}(-1)^n x^{2n+1}.$$

The finite geometric-series formula gives the remainder

$$\frac{x}{1+x^2}-S_N(x)=\frac{x(-x^2)^{N+1}}{1+x^2}.$$

For $0\le x\le1$, its absolute value is at most $x^{2N+3}$. Thus the absolute error in the integral is at most

$$\int_0^1x^{2N+3}\,dx=\frac{1}{2N+4},$$

which tends to zero. This justifies taking the limit of the integrated partial sums. Now use the alternating harmonic series:

$$I=\frac12\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1} =\frac12\ln 2.$$

As a check, direct integration gives $\int x/(1+x^2)\,dx=\tfrac12\ln(1+x^2)$, which also evaluates to $\tfrac12\ln2$ on these limits.

Convergence: what to check

A power series centered at $a$ has a radius of convergence $R$. It converges absolutely for $|x-a|<R$, diverges for $|x-a|>R$, and its behavior at $|x-a|=R$ must be checked separately.

If your entire interval is inside the radius—for example, it is a closed interval strictly contained in $|x-a|<R$—the series converges uniformly there, which permits term-by-term integration. If an endpoint lies exactly on the boundary, convergence of the series at that endpoint alone does not always settle the integration question. Use a remainder estimate, a suitable convergence theorem, or first integrate on a smaller interval and then justify taking a limit.

Common mistakes

• Using the expansion outside its range. The geometric expansion for $1/(1-u)$ is not valid as an infinite series when $|u|\ge1$.
• Ignoring endpoints. A series may converge at an endpoint without converging uniformly on the whole closed interval. Check the endpoint case before interchanging the sum and integral.
• Integrating a series that does not represent the function. Confirm the identity and its interval of validity, not just the formal pattern of coefficients.
• Assuming the answer must be elementary. The integrated series may be the most useful exact form, even when it does not simplify to elementary functions. It can still provide accurate numerical approximations.

When this approach is useful

Series integration is often effective when direct antiderivatives are unavailable or cumbersome. For instance, expanding $e^{-x^2}$ gives a convergent series for its integral, even though that antiderivative is not an elementary function. It is also useful for finding approximations: integrate a finite Taylor polynomial, then estimate the error using a bound on the omitted terms. Choose an expansion that converges quickly over the interval; a valid series with slowly decreasing terms may be less practical than another expansion or a different method.