Finding $\int_0^\infty\frac{\sin^{2}x}{1+x^4}dx$ I am trying to evaluate $$\int_0^\infty\dfrac{\sin^{2}x}{1+x^4}dx$$ and I am stuck on how to start. 
I am thinking the first step would be to substitute $$\dfrac{(1-e^{2ix})+(1-e^{-2ix})}{4}$$ for $\sin^{2}x$. 
I think the singularities are at multiples of $\frac{\pi}{4}$ but I am not sure what contour to use or how to proceed from the initial substitution for $\sin^{2}x$. 
Any help would be appreciated. 
Thanks 
 A: We have $\sin^2 x = \frac{1-\cos(2x)}{2}$ hence:
$$ I = \frac{1}{4}\int_{\mathbb{R}}\frac{dx}{1+x^4}-\frac{1}{4}\,\text{Re}\left(\int_{\mathbb{R}}\frac{e^{2ix}}{1+x^4}\,dx\right)$$
and the two integrals now appearing can be computed with the residue theorem; we just need the residues in the zeroes of $1+x^4$ in the upper-half plane. By computing them we get:

$$ I = \frac{\pi}{4\sqrt{2}}+\frac{\pi}{4\sqrt{2}}\left(\cos\sqrt{2}+\sin\sqrt{2}\right)e^{-\sqrt{2}}.$$

A: That's a good start.  You also might want to notice a symmetry that lets you take the integral from $-\infty$ to $+\infty$.  Now you put the "residue-calculus" tag on the question, so presumably you're going to end up looking at
residues for an analytic function inside a closed contour.  It's reasonable to
look at a contour going from $-R$ to $+R$ on the real axis, and then back to $-R$ along a semicircle in the upper half plane.  That's good for $\exp(2iz)$ but not for $\exp(-2iz)$ (why?).  But again, a symmetry will come to your rescue...
A: Try to remain in the domaine of reals with this. 
 
