complex integration of $\int_{-\infty}^{\infty}\sin^2(2x)/(1+x^2)$ I need to show $$\int_{-\infty}^{\infty}\sin^2(2x)/(1+x^2)=\pi sin ^2 2i$$
taking upper half circle as contour. I need to solve by residue theorem.
I tried putting $$f(z) = (e^{2iz})^2/ (1+z^2)  ........(1)$$  and then applying residue theorem,and equating imaginary parts but couldn't get the result.
Another obvious way is to simply take $$f(z)= \sin^2(2z)/(1+z^2)$$ and apply residue theorem. This gives result but this must be erroneous,isn't it? Because in case of sin/cos function, i know, we have to put $e^{iz}$ and then equate real/imaginary parts.
can anyone please help? how can I use equation (1) to arrive at the result??
 A: Note that $\pi\sin^{2}\left(2i\right)$ is not the right result. You could simplify a bit the integrand. Using some simple trigonometric identities we have $$I=\int_{-\infty}^{\infty}\frac{\sin^{2}\left(2x\right)}{1+x^{2}}dx=\frac{1}{2}\int_{-\infty}^{\infty}\frac{1}{1+x^{2}}dx-\frac{1}{2}\int_{-\infty}^{\infty}\frac{\cos\left(4x\right)}{1+x^{2}}dx$$ and $$\frac{1}{2}\int_{-\infty}^{\infty}\frac{1}{1+x^{2}}dx=\frac{\pi}{2}.$$ Now we have to study $$\frac{1}{2}\int_{-\infty}^{\infty}\frac{\cos\left(4x\right)}{1+x^{2}}dx.$$ Let us consider $$f\left(z\right)=\frac{1}{2}\frac{e^{4iz}}{z^{2}+1}$$ and take the upper half circle as contour. It is not difficult to see that the integral vanish over the semicirle as the radius $R$ goes to infinity. So $$\frac{1}{2}\int_{-\infty}^{\infty}\frac{e^{4iz}}{z^{2}+1}dz=\pi i\textrm{Res}_{z=i}\left(\frac{e^{4iz}}{z^{2}+1}\right)=\frac{\pi}{2e^{4}}$$ and obviously $$\int_{-\infty}^{\infty}\frac{\cos\left(4x\right)}{1+x^{2}}dx=\textrm{Re}\left(\int_{-\infty}^{\infty}\frac{e^{4iz}}{z^{2}+1}dz\right)$$ so $$I=\color{red}{\frac{\pi}{2}-\frac{\pi}{2e^{4}}}=\color{blue}{\frac{\pi\sinh\left(2\right)}{e^{2}}}$$ which is the right result.
