$\frac{1}{a^2+x^2}\ast \frac{1}{a^2+x^2}=\frac{2\pi /a}{4a^2+x^2} (a>0)$

Question

I.e. $\displaystyle\int_{\mathbb{R}} \frac{1}{a^2+(x-y)^2}\cdot\frac{1}{a^2+y^2} dy=\frac{2\pi /a}{4a^2+x^2}$. If anyone feels like giving me a brief hint about how to get started on this I'd be grateful.

Answer 1

A standard technique to attack this is through Fourier analysis.

Consider the function $f$ defined by $$f(s)=\frac{1}{a^2+s^2}$$ We wish to show $$(f*f)(x) = \frac{2\pi/a}{4a^2 +x^2} = \frac{2\pi/a}{4(a^2 +(x/2)^2)} = \frac{\pi}{2a}f(x/2) $$

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A note on the Fourier transform...

There are different choices of the Fourier transform (the taste of choice depends in a sense on "where to put $\pi$"), among one is $$\hat{g}(t)=\int g(x)e^{-itx}dx$$ which we use from here on. Then one nice property is that convolution is transformed into ordinary multiplication: $$\widehat{(f*g)}(x)=\hat{f}(t)\hat{g}(t)$$ We also have the inverse Fourier transform that says $$\hat{\hat{f}}(t)=2\pi \check{f}(t)= 2\pi f(-t)$$

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Now I will leave a problem that will resolve the main problem using the above comment on the Fourier transform.

Problem: Put $E(b) = e^{-a|b|} $ what is $\hat{E}$?

Answer 2

If I am understanding the problem correctly, you want to integrate the slightly complicated function of $x$ and $y$ from $y=-\infty$ to $y=\infty$. It is doable, but unpleasant.

As usual with this sort of rational function (of $y$), we can use partial fractions. So try to express your integrand as $$\frac{Py+Q}{a^2+(x-y)^2} +\frac{Ry+S}{a^2+y^2}.$$ Use your favourite partial fractions procedure, say cross-multiplying and using the fact that the numerator must be identically $1$ to find $P,Q,R,S$. The only difference from the ordinary first integration course is that $P,Q,R,S$ will be functions of $x$. The calculation was not much fun, but less terrible than I feared. I get: $$P=-\frac{2}{x(4a^2+x^2)}\qquad R=-\frac{2}{x(4a^2+x^2)},$$ $$Q=\frac{3}{4a^2+x^2}\qquad S=\frac{1}{4a^2+x^2}.$$ The rest should be familiar. When you are doing the integrating, you will get $\log$ terms and $\arctan$ terms. You need to be careful with the $\log$ part, keep the two $\log$ calculations together, since I think individually their integrals blow up, but the difference doesn't.

Running out of time for today! But I assure you that despite the apparent mess, the rest is not as bad as it looks. The $4a^2+x^2$ is in the denominator everywhere, so forget about it and remember later.

Final suggestion: When you are integrating $(Py+Q)/(a^2+(x-y)^2)$, rewrite as $(P(y-x) +Q +x)/(a^2+(x-y)^2)$. For the $P(y-x)/(a^2+(x-y)^2)$ part, make the change of variable $x-y=u$ and something very nice will happen.