<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] Here using Fourier transform may simplify things considerably:
Let $\phi(x)=\frac{1}{1+x^2}$, and $\phi_a(x)=\frac{1}{a}\phi(a^{-1}x)=\frac{a}{a^2+x^2}$.
Recall that $\pi e^{-2\pi|t|}=\int e^{-2\pi itx}\frac{1}{1+x^2}\,dx=\widehat{\phi_1}(t)$. Hence
$$\widehat{\phi_a}(t)=\frac{1}{a}\int e^{-2\pi itx}\phi(xa^{-1})\,dx=\frac{1}{a}\int e^{2\pi ia t(a^{-1}x)}\phi(a^{-1}x)\,dx=\widehat{\phi_1}(at)$$
Thus
$$\widehat{\phi_a*\phi_a}(t)=(\widehat{\phi_a}(t))^2=(\widehat{\phi_1}(at))^2=\pi^2e^{-4\pi a|t|}$$
That is,
$$
I_a=a^{-2}(\phi_a*\phi_a)(x)=a^{-2}\pi^2\mathcal{F}^{-1}(e^{-4\pi a|t|})(x)=\frac{2\pi}{a(a^2+x^2)}$$
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