<s>[INST] Surely You're Joking, Mr. Feynman! $\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}\,dx$ 
Prove the following
\begin{equation}\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}\,dx=\frac{\pi}{4}+\frac{\pi}{4e^2}\end{equation}

I would love to see how Mathematics SE users prove the integral preferably with the Feynman way (other methods are welcome). Thank you. (>‿◠)✌

Original question:

And of course, for the sadist with a background in differential
equations, I invite you to try your luck with the last integral of the
group.
\begin{equation}\int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}\,dx\end{equation}

Source: Integration: The Feynman Way
 [/INST] This integral is readily evaluated using Parseval's theorem for Fourier transforms.  (I am certain that Feynman had this theorem in his tool belt.)  Recall that, for transform pairs $f(x)$ and $F(k)$, and $g(x)$ and $G(k)$, the theorem states that
$$\int_{-\infty}^{\infty} dx \, f(x) g^*(x) = \frac1{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G^*(k) $$
In this case, $f(x) = \frac{\sin^2{x}}{x^2}$ and $g(x) = 1/(1+x^2)$.  Then $F(k) = \pi (1-|k|/2) \theta(2-|k|)$ and $G(k) = \pi \, e^{-|k|}$. ($\theta$ is the Heaviside function, $1$ when its argument is positive, $0$ when negative.)  Using the symmetry of the integrand, we may conclude that
$$\begin{align}\int_0^{\infty} dx \frac{\sin^2{x}}{x^2 (1+x^2)} &= \frac{\pi}{2} \int_0^{2} dk \, \left ( 1-\frac{k}{2} \right ) e^{-k} \\ &= \frac{\pi}{2} \left (1-\frac1{e^2} \right ) - \frac{\pi}{4} \int_0^{2} dk \, k \, e^{-k} \\ &= \frac{\pi}{2} \left (1-\frac1{e^2} \right ) + \frac{\pi}{2 e^2}  - \frac{\pi}{4} \left (1-\frac1{e^2} \right )\\ &= \frac{\pi}{4} \left (1+\frac1{e^2} \right )\end{align} $$
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