Problem

Compute the integral

$$ I = \int_0^\infty \frac 1 {1+x^4} dx. $$

Answer

This seems to be a popular problem illustrating various mathematical techniques. The standard approach is to use complex analysis and the Cauchy method of residues. The problem may be solved with more elementary techniques by using a series of clever substitutions. Both approaches rely on quite lengthy computations to obtain the answer

$$ I = \pi \frac {\sqrt 2} 4. $$

Here, we would like to show that the answer may be obtained quite easily in the form of power series.

Power series solution

We write

$$ I = \int_0^1 \frac 1 {1+x^4} dx + \int_1^\infty \frac 1 {1+x^4} dx = I_1 + I_2. $$

For the first integral we have

$$ I_1 = \int_0^1 \sum_{k=0}^\infty (-1)^k x^{4k} dx = \sum_{k=0}^\infty (-1)^k \frac 1 {4k + 1}. $$

For the second integral we make the substitution $x = 1/y$ and get

$$ I_2 = \int_0^1 \frac {y^2} {1+y^4} dy = \int_0^1 \sum_{k=0}^\infty (-1)^k y^{4k + 2} dy = \sum_{k=0}^\infty (-1)^k \frac 1 {4k + 3}. $$

So the final result is

$$ I = \sum_{k=0}^\infty (-1)^k \frac 1 {4k + 1} + \sum_{k=0}^\infty (-1)^k \frac 1 {4k + 3}. $$

For example, if we take the first few members of each series we obtain the approximation

$$ I \approx \left(1 - \frac 1 5 + \frac 1 9 - \frac 1 {13} + \frac 1 {17}\right) + \left(\frac 1 3 - \frac 1 7 + \frac 1 {11} - \frac 1 {15}\right) \approx 1.1077, $$

whose error is less than $0.003$ of the exact value of $I$.

It would be an interesting exercise to find the sums of the above power series in closed form.