<s>[INST] Simplify convolution integral $\int_{-\infty}^\infty\frac{1}{(a^2+t^2)(a^2+(x-t)^2) }dt$ I am trying to show that $$\frac{1}{a^2+x^2} \ast \frac{1}{a^2+x^2}=\int_{-\infty}^\infty\frac{1}{(a^2+t^2)(a^2+(x-t)^2) }dt = \frac{2\pi}{a(4a^2+x^2)}$$  I wrote out the integral for the convolution and it becomes a big exercise in partial fractions and integration of rational functions.
$$ \int(\frac{1}{a^2+t^2})  (\frac{1}{a^2+(x-t)^2}) dt=
\int\frac{x + 2 t}{x (4 a^2 + x^2) (a^2 + t^2)} + \frac{3 x - 2 t}{x (4 a^2 + x^2) (a^2 + x^2 - 2 xt + t^2)} dt =$$
$$=\frac{a (log(a^2 + t^2) - log(a^2 + (x - t)^2)) + x \cdot tan^{-1}(\frac{t}{a}) + x \cdot tan^{-1}(\frac{t - x}{a})}
{a x (4 a^2 + x^2)}
$$
Plugging in the bounds of plus/minus infinity makes the logs cancel out, and gives the 2$\pi x$ from the arctan terms.
My question: Is there a way to simplify / make this any easier using the properties of the convolution?  Or are these problems just 'plug and chug' like back in my engineering days?
 [/INST] Symmetrize the integrand with $y= t-\frac x2$
\begin{align}
& \int_{-\infty}^\infty\frac{1}{(a^2+t^2)(a^2+(x-t)^2) }dt\\
= & \int_{-\infty}^\infty\frac{1}{(a^2+(y+\frac x2)^2)(a^2+(y-\frac x2)^2)}dy\\
=&\ 
\frac2{x(4a^2+x^2)}\int_{-\infty}^\infty
 \frac{{\frac x2 +(y+\frac x2)}}{a^2+(y+\frac x2)^2 }
+ \frac{{\frac x2 -(y-\frac x2)}}{a^2+(y-\frac x2)^2 } \overset{y\pm\frac x2\to y}{dy}\\
= &\ \frac2{4a^2+x^2}\int_{-\infty}^\infty\frac{1}{ a^2+y^2}dy
= \frac{2\pi}{a(4a^2+x^2)}
\end{align}
</s>