From a Rational Function to an Infinite Sum
A rational integrand can sometimes be written as a geometric series whose terms are easy to integrate. The main opportunity is that integration changes each power into a simple coefficient, so one definite integral can produce an exact identity for an infinite sum.
There is an essential condition: the series representation must be valid across the interval of integration in a way that permits term-by-term integration. A formal substitution into the geometric-series formula is not enough. We will make the interval check explicit, justify the interchange, and then apply the method to several integrals.
Geometric Expansions on an Interval
For a number \(u\) with \(|u|<1\), the geometric series gives
For an integrand, \(u\) usually depends on \(x\). Suppose \(u\) is continuous on a closed interval \([a,b]\), and there is a number \(\rho<1\) such that \(|u(x)|\leq \rho\) for every \(x\in[a,b]\). Then
throughout that interval. The strict uniform bound is important: it ensures that the geometric remainders become small throughout the entire interval, not merely at individual points.
Why Term-by-Term Integration Is Valid
The justification can be seen directly from the finite geometric-sum formula. For every nonnegative integer \(N\),
If \(|u(x)|\leq\rho<1\), the absolute value of the remainder is at most \(\rho^{N+1}/(1-\rho)\), uniformly over the interval. This bound tends to zero as \(N\) tends to infinity. Consequently, the integrals of the finite sums converge to the integral of the function.
The same conclusion holds after multiplying by a constant. It also applies term by term to finite sums of such geometric expansions. The argument does not say that every pointwise-convergent series can be integrated term by term; the uniform remainder bound is what justifies this method.
Rewrite the denominator so the integrand contains \(1/(1-u(x))\), possibly with a constant factor.
Find a bound \(|u(x)|\leq\rho<1\) that holds for every point of the interval.
Use the theorem to integrate each power and form the resulting infinite sum.
When possible, evaluate the original integral directly and compare it with the series identity.
Worked Example: A Geometric Series for an Arctangent Integral
Worked Example: A Geometric Series for an Arctangent Integral
Evaluate \(\displaystyle \int_0^{1/2}\frac{1}{1+x^2}\,dx\) and express its value as a series.
Write the denominator in geometric form by taking \(u(x)=-x^2\). For \(0\leq x\leq 1/2\), \(|u(x)|=x^2\leq1/4<1\). Thus the expansion is uniform on this interval:
Integrate term by term. For each \(n\), the power rule gives \(\int_0^{1/2}x^{2n}\,dx=(1/2)^{2n+1}/(2n+1)\). Therefore,
To check the result directly, an antiderivative is \(\arctan x\), since \(\frac{d}{dx}\arctan x=1/(1+x^2)\). Hence the integral equals \(\arctan(1/2)-\arctan(0)=\arctan(1/2)\). The resulting identity is
The first terms are \(1/2-1/24+1/160-\cdots\). The direct antiderivative confirms the value, and the uniform convergence check justifies the series evaluation.
Worked Example: A Related Integral Gives a Different Sum
Worked Example: A Related Integral Gives a Different Sum
Evaluate \(\displaystyle \int_0^{1/2}\frac{x^2}{1+x^2}\,dx\) as a series and check it directly.
Use the same geometric expansion as before. Multiplication by \(x^2\) gives
The bound \(|x^2|\leq1/4<1\) on \([0,1/2]\) still guarantees uniform convergence. Integrating each term gives
For a direct check, rewrite the integrand as \(1-1/(1+x^2)\). An antiderivative is \(x-\arctan x\), because its derivative is \(1-1/(1+x^2)=x^2/(1+x^2)\). Evaluating at the endpoints gives
Combining the direct evaluation with the series calculation yields
This check also follows by subtracting the first example’s integral from \(\int_0^{1/2}1\,dx=1/2\), in agreement with the rewritten integrand.
Weighted Geometric Series
Some rational functions produce coefficients that depend on \(n\), rather than just a constant multiple of \(u(x)^n\). A useful example comes from differentiating the geometric-series identity inside its interval of convergence:
On any closed interval contained strictly inside \((-1,1)\), this differentiated power series converges uniformly. The factor \(n+1\) does not prevent integration there: the geometric decay of \(|x|^n\) dominates that factor. The next example uses this representation.
Worked Example: Integrating a Weighted Geometric Series
Worked Example: Integrating a Weighted Geometric Series
Evaluate \(\displaystyle \int_0^{1/2}\frac{1}{(1+x)^2}\,dx\) by a series and verify the answer.
Since \(0\leq x\leq1/2<1\), the weighted geometric series converges uniformly on the interval. Integrating term by term gives
The last series is geometric with first term \(1/2\) and ratio \(-1/2\). Its sum is
For the direct check, an antiderivative is \(-1/(1+x)\), whose derivative is \(1/(1+x)^2\). Therefore,
The direct value agrees with the sum obtained by integrating the series.
A Common Mistake: Ignoring the Interval
A geometric expansion may be valid near one point but invalid across the entire interval of integration. For example, the identity
requires \(|x|<1\). It is therefore incorrect to use it term by term on \([0,2]\). That would produce the proposed expression
The terms of this series do not approach zero: their absolute values \(2^{2n+1}/(2n+1)\) grow without bound. Thus the expression cannot be the value of the integral. In fact, the original integrand is continuous on \([0,2]\), so its definite integral is finite. The error was not in integrating a power; it was using a series outside the interval where its expansion applies.
Why This Method Matters
The method turns an integral into a systematic way of producing exact sums. Each term is integrated using familiar power rules, while a direct evaluation of the integral can identify the value of the resulting series. The interval check is not a technical afterthought: it connects the algebraic expansion to the function being integrated and ensures that the sum actually represents the integral.
When applying the method, keep three questions separate: Is the algebraic geometric expansion correct? Does it converge uniformly on the chosen interval? And does the term-by-term integral agree with a direct evaluation or another valid calculation? Answering all three makes the resulting sum reliable.
Check Your Understanding
For each question, check both the series calculation and the interval condition.
- On \([0,1/3]\), what geometric expansion represents \(1/(1+x^2)\), and what series results from integrating it?
- Why does a bound \(|u(x)|\leq\rho<1\) on a closed interval justify integrating \(\sum_{n=0}^{\infty}u(x)^n\) term by term?
- What sum results from integrating the series for \(x/(1+x^2)\) on \([0,1/2]\)?
- Why is the expansion \(\sum_{n=0}^{\infty}(-1)^n x^{2n}\) unsuitable for direct term-by-term integration on \([0,2]\)?
- For the weighted series for \(1/(1+x)^2\), which factor in each term cancels when the term is integrated?
