1

I am stuck computing a convolution because I need to be able to solve it using only the definition of the concept. I need to compute $(f*f)(x)$, where $$f(x) = \frac{1}{x^2+1}$$ So far I've only been able to do this:

$$(f*f)(x) = \int_{-\infty}^\infty f(y)\cdot f(x-y)dy = \int_{-\infty}^\infty \frac{1}{y^2-1}\frac{1}{(x-y)^2+1}dy$$ Expanding the parentheses in the denominator and solving the quadrating equation on $y$ returns the roots $y = x\pm i$, and therefore: $$(f*f)(x) = \int_{-\infty}^\infty \frac{1}{y+1}\frac{1}{y-1}\frac{1}{y-x-i}\frac{1}{y-x+i}dy$$ But I don't know how to keep going to use the residue theorem. How could I transform this into an intgral over a closed path? Any help would be appreciated.

José Carlos Santos
  • 427,504
coffee_pls
  • 147

1 Answers1

2

What you should compute is$$\int_{-\infty}^\infty\frac1{y^2+1}\frac1{(x-y)^2+1}\,\mathrm dy.\tag1$$If$$f(z)=\frac1{z^2+1}\frac1{(x-z)^2+1},$$then$$\operatorname{res}_{z=i}f(z)=\frac i{2x(x-2i)}\quad\text{and}\quad\operatorname{res}_{z=x+i}f(z)=-\frac i{2x(x+2i)}.$$So, $(1)$ is equal to$$2\pi i\left(\frac i{2x(x-2i)}-\frac i{2x(x+2i)}\right)=\frac{2\pi}{x^2+4}.$$

José Carlos Santos
  • 427,504
  • Why are you only computing the residues at two of the poles of $f(z)$? Aren't $z=-i$ and $z=x-i$ also poles? Sorry, I do not understand that part. – coffee_pls Dec 20 '21 at 08:59
  • Because I am using a standard method for computing such integrals: I see it as the limit, as $R$ goes to $\infty$, of the integral along $[-R,R]$ followed by the integral along the half-circle centered at $0$ that goes from $R$ to $-R$ along the upper half-plane. So, only the points with positive imaginary part matters. See this question. – José Carlos Santos Dec 20 '21 at 09:06
  • Okay, this makes a lot of sense. I wasn't aware about that method for solving these types of integrals. Thanks for such a clear answer! – coffee_pls Dec 20 '21 at 09:19