Evaluating $\int_0^\infty \frac{dx}{1+x^4}$. 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?

You could mess about with integration by substitution methods. 
For instance, if you change the expression to this: 
$$\frac{1}{2}\int_0^\infty \frac{(x^2+1)-(x^2-1)}{x^4+1}dx = \frac{1}{2}\int_0^\infty \frac{(x^2+1)}{x^4 + 1}-\frac{(x^2-1)}{x^4+1}dx$$
which you can then separate out as:
$$\frac{1}{2}\left[\int_0^\infty \frac{(x^2+1)}{x^4+1}dx - \int_0^\infty \frac{(x^2-1)}{x^4+1}dx\right]$$
and then divide the top and bottom in both expressions by $x^2$ to get: 
$$\frac{1}{2}\left[\int_0^\infty \frac{1+ \frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx - \int_0^\infty \frac{1- \frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx\right]$$
and then you can change some variables. If you take:
$u= x- \frac{1}{x}$ so $du= 1 + \frac{1}{x^2}$ and $v= x+\frac{1}{x}$ and $dv = 1 - \frac{1}{x^2}$ 
you can square the functions: 
$u^2 = x^2 -2 +\frac{1}{x^2}$ or $u^2 + 2 = x^2+\frac{1}{x^2}$
$v^2 = x^2 +2 +\frac{1}{x^2}$ or $v^2-2  = x^2 +\frac{1}{x^2}$
and when you substitute in you end up with 
$$\frac{1}{2}\left[\int_0^\infty \frac{du}{u^2+2} - \int_0^\infty \frac{dv}{v^2-2}\right]$$
and from there it should be easier to do, no? I think this was the way I approached it in a class last semester and I saw someone else in the interwebs had done something similar.  :-) so I could be wrong. But I think this should work. Or you could I think rearrange the above substitutions into an integration by parts for each expression. But when I did that it was a long process. 
