Simplify convolution integral $\int_{-\infty}^\infty\frac{1}{(a^2+t^2)(a^2+(x-t)^2) }dt$

Question

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?

Answer 1

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, $$ I_a=a^{-2}(\phi_a*\phi_a)(x)=a^{-2}\pi^2\mathcal{F}^{-1}(e^{-4\pi a|t|})(x)=\frac{2\pi}{a(a^2+x^2)}$$

Answer 2

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}

Answer 3

Less elegant than @Quanto's answer, let us write $$a^2+(x-t)^2=(t-r)(t-r)$$ with $r=(x+ia)$ and $r=(x-ia)$ and use partial fraction decomposition $$\frac{1}{(a^2+t^2)(a^2+(x-t)^2) }=\frac{r s+r t+s t-a^2}{\left(a^2+r^2\right) \left(a^2+s^2\right) \left(t^2+a^2\right)}+$$ $$\frac{1}{\left(a^2+r^2\right) (r-s) (t-r)}-\frac{1}{\left(a^2+s^2\right) (r-s) (t-s)}$$ Three simple antiderivatives and then the result.