TITLE: Evaluating $\int_0^\infty \frac{dx}{1+x^4}$.
QUESTION [13 upvotes]: Can anyone give me a hint to evaluate this integral?
$$\int_0^\infty \frac{dx}{1+x^4}$$
I know it will involve the gamma function, but how?

REPLY [2 votes]: This was one of the questions in my final Infinitesimal Calculus I course (one year long course first undergraduate year)...and thus one has no idea whats' going on and must then try more or less simple things, so partial fractions:
$$x^4+1=(x^2+\sqrt2\,x+1)(x^2-\sqrt2\,x+1)\implies$$
$$\frac1{x^4+1}=\frac{Ax+B}{x^2+\sqrt2\,x+1}+\frac{Cx+D}{x^2-\sqrt2\,x+1}\;,\;A,B,C,D\in\Bbb R\implies$$
$$1=(Ax+B)(x^2-\sqrt2\,x+1)+(Cx+D)(x^2+\sqrt2\,x+1)$$
Comparing respective coefficients of $\,x\,$  and checking what happens with $\,x=0\,$ and etc., we get
$$\frac1{x^4+1}=\frac{x+\sqrt2}{2\sqrt2(x^2+\sqrt2\,x+1)}-\frac{x-\sqrt2}{2\sqrt2(x^2-\sqrt2\,x+1)}$$
and
$$\int\frac{x+\sqrt2}{x^2+\sqrt2\,x+1}dx=\frac12\int\left(\frac{2x+\sqrt2}{x^2+\sqrt2\,x+1}+\frac{\sqrt2-\frac1{\sqrt2}}{x^2+\sqrt2\,x+1}\right)dx=$$
$$=\frac12\log\left(x^2+\sqrt2\,x+1\right)+\frac1{2\sqrt2}\int\frac{dx}{\left(x+\frac1{\sqrt2}\right)^2+\frac12}=$$
$$=\frac12\log\left(x^2+\sqrt2\,x+1\right)+\frac1{\sqrt2}\int\frac{dx}{1+\left(\frac x{\sqrt2}+\frac12\right)^2}=$$
$$=\frac12\log\left(x^2+\sqrt2\,x+1\right)+\arctan\left(\frac x{\sqrt2}+\frac12\right)$$
Well, take it from here also with the other integral, put limits and etc.