Could anybody check this integral? in a lengthy calculation by hand I got that
$$ \frac{2}{\pi \sigma_k} \int_{-\infty}^{\infty} \frac{sin^2(\frac{\sigma_k}{2}(v_gt-x))}{(v_gt-x)^2} dx =1$$
Now I was wondering whether there is anybody who could check this ( with a CAS or by hand )? If you decide to do the last option, please note that:
$$  \int_{-\infty}^{\infty} \frac{sin^2(x)}{x^2} dx = \pi$$
Thanks in advance
 A: It takes one line with the change of variable $y=\frac{\sigma_k}{2}(v_gt-x)$
$$ \frac{2}{\pi \sigma_k} \int_{-\infty}^{\infty} \frac{\sin^2(\frac{\sigma_k}{2}(v_gt-x))}{(v_gt-x)^2} dx = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{\sin^2y}{y^2} dy =1  $$
The detail of the change of variable goes as follows:
$$y=\frac{\sigma_k}{2}(v_gt-x) \implies dy = -\frac{\sigma_k}{2}dx$$
With $y$ going from $\infty$ to $-\infty$ due to the difference of sign. We rework the initial integral to get the form that exhibits $y$ explicitly.
$$ \frac{2}{\pi \sigma_k} \int_{-\infty}^{\infty} \frac{\sin^2(\frac{\sigma_k}{2}(v_gt-x))}{(v_gt-x)^2} dx=\frac{\sigma_k}{2\pi } \int_{-\infty}^{\infty} \frac{\sin^2(\frac{\sigma_k}{2}(v_gt-x))}{\left(\frac{\sigma_k}{2}(v_gt-x)\right)^2} dx $$
Then substituting for $y$ and changing the limits accordingly, we get:
$$= -\frac{1}{\pi} \int_{\infty}^{-\infty} \frac{\sin^2y}{y^2} dy = \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{\sin^2y}{y^2} dy  =1$$
A: Let's try together. Assume $\sigma > 0$. Substitute $u = \frac{\sigma}{2} (vt-x)$ to get $du = -\sigma dx/2$ and $[-\infty,\infty] \to [\infty,-\infty]$ we have
$$
\int_{-\infty}^\infty \frac{\sin^2 \left(\frac{\sigma}{2} (vt-x) \right)}{(vt-x)^2} dx
 = \int_\infty^{-\infty} \frac{\sin^2 u}{\frac{4}{\sigma^2}u^2} \frac{-2du}{\sigma}
 = \frac{-2}{\sigma} \frac{\sigma^2}{4} (-\pi)
 = \frac{\sigma \pi}{2},
$$
so sounds like you did it correctly.
A: $$J=A\int_{-\infty}^{+\infty}\frac{\sin^2(k(a-x)}{(a-x)^2}dx=Ak\pi$$
Because you have $k=\frac{1}{A}$, $J=1$
