Integral with residues $\int_0^\infty \tfrac{\sin^2(x)}{1+x^4}dx$ I am trying to calculate
$\displaystyle\int_0^\infty \dfrac{\sin^2(x)}{1+x^4}dx$
using method of residues. I have already seen this post, "Integrating $\int_{-\infty}^\infty \frac{1}{1 + x^4}dx$ with the residue theorem" And am integrating a quarter circle in the complex plane with a simple pole at $\exp(\frac{i\pi}{4})$. So if I put $z=e^{i\theta}, \sin^2(x)=-\frac{1}{4}(z-\frac{1}{z})$. Even so the residues are giving me difficulty. Any help would be great. 
 A: First note that the integrand is even, so
$$
\int_{0}^\infty \frac{\sin^2 x}{1+x^4}\,dx = 
\frac12 \int_{-\infty}^\infty \frac{\sin^2 x}{1+x^4}\,dx.
$$
Furthermore $\sin^2 x = \dfrac{1-\cos 2x}{2}$ so
$$
\int_{0}^\infty \frac{\sin^2 x}{1+x^4}\,dx = 
\frac12 \int_{-\infty}^\infty \frac{\sin^2 x}{1+x^4}\,dx =
\frac14 \int_{-\infty}^\infty \frac{1-\cos 2x}{1+x^4}\,dx.
$$
Define a function
$$
f(z) = \frac{1-e^{2iz}}{1+z^4}.
$$
Then $\operatorname{Re}(f(x)) = \dfrac{1-\cos 2x}{1+x^4}$ for $x$ real. Let's compute $\int_C f(z)\,dz$ over the boundary of a large semi-disc: $C = \partial\{ z = x+iy : |z| < R, y = 0 \}$.
On the semi-circle (by the triangle inequality), we get
$$
|f(z)| \le \frac{1 + |e^{2iz}|}{|z|^4-1} \le \frac{2}{R^4-1}
$$
since $|e^{2i(x+iy)}| = |e^{-2y}| \le 1$ for $y \ge 0$. 
The standard estimation lemma (ML-inequality) shows that
$$
\left| \int_{C_R^+} f(z)\,dz \right| \le \pi R \cdot \frac{2}{R^4-1} \to 0
$$
as $R \to \infty$. (Here $C_R^+$ is the semi-circle.)
Finally, the residue theorem shows that
\begin{align}
\int_C f(z)\,dz &= 2\pi i \big( \operatorname{Res}\limits_{\exp(i\pi/4)}(f(z)) + 
\operatorname{Res}\limits_{\exp(3i\pi/4)}(f(z)) \big) \\
&= \frac{\pi\sqrt 2}{2}\,\big( 1- \exp(-\sqrt 2)(\sin \sqrt 2 + \cos \sqrt 2) \big).
\end{align}
(Tedious algebra omitted.)
Putting everything together, and taking the real part we get $1/4$ of the above, i.e.
$$
\int_{0}^\infty \frac{\sin^2 x}{1+x^4}\,dx = \frac{\pi\sqrt 2}{8}\,\big( 1- \exp(-\sqrt 2)(\sin \sqrt 2 + \cos \sqrt 2) \big).
$$
A: Notice that your function, with $z$ in place of $x$, is symmetric in both the real and imaginary axis.
$$f(z)=\frac{sin^2(z)}{1+z^4} \longrightarrow f(x+iy) = f(-x+iy) = f(x-iy) = f(-x-iy)$$
With this in mind, try the following contours.

If you add them up, you the top and bottom should cancel (due to opposing directions of orientation), and the two horizontal strips should give you four times the value you are looking for.
