Differentiation wrt parameter $\int_0^\infty \sin^2(x)\cdot(x^2(x^2+1))^{-1}dx$

Consider for $a>0$ $$ I(a)=\int_0^\infty\frac{\sin^2(ax)}{x^2(x^2+1)}dx $$ Differentiate it twice. Since $$ \int_0^\infty\frac{\cos(kx)}{x^2+1}dx=\frac{\pi}{2e^k} $$ for $k>0$ we get $I''(a)=\pi e^{-2a}$. Note that $I'(0)=I(0)=0$, so after solving respective IVP we get $$ I(a)=\frac{\pi}{4}(-1+2a+e^{-2a}) $$ It is remains to substitute $a=1$.

Well, as an alternative (and I promise, none of the dreaded complex analysis stuff), we could use Parseval's theorem for Fourier transforms:

For example, the FT of $(\sin{x}/x)^2$ is

$$\int_{-\infty}^{\infty} dx \: \frac{\sin^2{x}}{x^2} e^{i k x} = \begin{cases} \\\pi \left (1 - \frac{|k|}{2} \right ) & |k| \le 2 \\ 0 & |k| > 2 \end{cases}$$

The FT of $1/(1+x^2)$ is

$$\int_{-\infty}^{\infty} dx \: \frac1{1+x^2} e^{i k x} = \pi \, e^{-|k|}$$

By Parseval's theorem,

$$\begin{align}\int_0^{\infty} dx \: \frac{\sin^2{x}}{x^2} \frac1{1+x^2} &= \frac{\pi}{2} \int_0^2 dk \, \left (1 - \frac{k}{2} \right ) e^{-k}\\ &= \frac{\pi}{2} \left (1-e^{-2}- \frac12 (1-3 e^{-2}) \right )\\ &= \frac{\pi}{4} \left (1+\frac1{e^2} \right )\end{align}$$

Well, as an alternative (like Mr. Ron Gordon did). \begin{align} \int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}dx&=\int_0^\infty\left[\frac{\sin^2x}{x^2}-\frac{\sin^2x}{1+x^2}\right]dx\\ &=\int_0^\infty\frac{\sin^2x}{x^2}dx-\frac{1}{2}\int_0^\infty\frac{1-\cos2x}{1+x^2}dx\\ &=\frac{\pi}{2}-\frac{1}{2}\int_0^\infty\frac{1}{1+x^2}dx+\frac{1}{2}\int_0^\infty\frac{\cos2x}{1+x^2}dx\\ &=\frac{\pi}{2}-\frac{1}{2}\frac{\pi}{2}+\frac{1}{2}\frac{\pi}{2e^2}\\ &=\frac{\pi}{4}+\frac{\pi}{4e^2} \end{align} where I use these links: $\displaystyle\int_0^\infty\frac{\sin^2x}{x^2}dx$ and $\displaystyle\int_0^\infty\frac{\cos2x}{1+x^2}dx$ to help me out.

Unfortunately, this is not differentiation with respect to parameter method (the Feynman way) but I still love this method. (>‿◠)✌