Find the integral $\int\limits_{-\infty}^{+\infty}{\frac{dt}{(1+(x-t)^2)(1+t^2)}}$

Question

My problem is to calculate auto-convolution of $f(x) = \frac{1}{1+x^2}$. I know that $$(f \star f)(x) = \int\limits_{-\infty}^{+\infty}{\frac{dt}{(1+(x-t)^2)(1+t^2)}} = \frac{2 \pi}{x^2+4},$$ but I cannot solve it not using calculators. Thanks in advance for any advices.

Answer 1

We can use residue theorem to compute improper integral. Let $C$ be a union of upper-semicircle whose center is $0$ and radius is $R$ (sufficiently large), and segment from $-R$ to $R$. Then by residue theorem, \begin{align} \int_C {\frac{dz}{(1+(x-z)^2)(1+z^2)}}&=2\pi i (\operatorname{Res}(f;i)+\operatorname{Res}(f;x+i))\\ &=2\pi i\left(\frac{1}{2i(2i-x)(-x)}+\frac{1}{x(x+2i)2i}\right)\\ &=2\pi i \cdot \frac{1}{2i}\cdot \frac{2}{x^2+4}\\ &=\frac{2\pi}{x^2+4} \end{align} and semi-circle part integral goes to $0$ as $R\to\infty$.

Answer 2

Use the convolution theorem: the Fourier transform of a convolution of two functions $f$ and $g$ is equal to the product of the FT of the functions $F$ and $G$. Thus,

$$\int_{-\infty}^{\infty} dx \, (f * g)(x) \, e^{i k x} = 2 \pi \,F(k)G(k) $$

The factor of $2 \pi$ is a result of the scaling I chose. Therefore,

$$(f*g)(x) = \int_{-\infty}^{\infty} dk \, F(k) G(k) \, e^{-i k x} $$

In your case, $f(x) = g(x) = 1/(1+x^2)$. Thus, $F(k) = G(k) = \pi \,e^{-|k|}$, and

$$\begin{align}\int_{-\infty}^{\infty} \frac{dx'}{[1+(x-x')^2](1+x'^2)} &= \pi^2 \int_{-\infty}^{\infty} dk \, e^{-2 |k|} e^{-i k x}\\ &= \frac{\pi^2}{2} \int_{-\infty}^{\infty} dk \, e^{-|k|} e^{-i k x/2}\\ &= \frac{\pi}{2} \frac1{1+x^2/4} \\ &= \frac{2 \pi}{4+x^2}\end{align}$$