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

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.

-
@Paul: I'm not sure I understand your edit. – Tim kinsella Jan 28 '12 at 5:54
Tim, where did you came across this? In the title you used the symbol * which usually (in analysis) refer to convolution between two functions $f$ and $g$, and it is then defined by $f*g (x)=\int f(y)g(x-y)dy$ (in case $f$ and $g$ live on the reals). – AD. Jan 28 '12 at 7:03
Yes I intended $\ast$ to mean convolution. Have I used it incorrectly? The question comes from an exercise in the book "Lebesgue Integration on Euclidean Space" by Frank Jones. – Tim kinsella Jan 28 '12 at 7:30
Thanks Tim, perhaps the author said something in the foreword regarding the background, anyway it is good to have some Fourier theory in the luggage in order to see how useful the Lebesgue theory is (I would recommend the Fourier chapters in Rudin "Real & Complex analysis" or perhaps you would like "Fourier Series and Integrals" by Dyn and McKean? ) – AD. Jan 28 '12 at 21:28

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 $$(f*f)(x) = \frac{2\pi/a}{4a^2 +x^2} = \frac{2\pi/a}{4(a^2 +(x/2)^2)} = \frac{\pi}{2a}f(x/2)$$

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 $$\hat{\hat{f}}(t)=2\pi \check{f}(t)= 2\pi f(-t)$$

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 $\hat{E}$?

-
Thanks. I am about to learn about the Fourier transform so I will keep this in mind. It is always reassuring to learn that there is a more elegant solution on the horizon. – Tim kinsella Jan 28 '12 at 8:01
I see. $\hat{E}$ is something like $2af$. So now presumably I just need to plug in $\hat{E}/2a$ for $f$ in $f\ast f (x) =\frac{\pi}{2a}f(x/2)$ and use those properties of the transform to work backwards to something true. – Tim kinsella Jan 28 '12 at 8:40
Yes! That is the idea! – AD. Jan 28 '12 at 21:28

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.

-
Goodness! I would have settled for "integrate by parts." Thanks Andre. – Tim kinsella Jan 28 '12 at 7:28
It's not really parts, it is substitution. Basically you will be working after a while with $z/(a^2+z^2)$ (let $w=a^2+z^2)$ and $1/(a^2+z^2)$ (let $z=\tan\theta$). – André Nicolas Jan 28 '12 at 7:33
yeah, sorry i meant "use partial fractions." Both have "part" i guess and its been a while since calculus :) – Tim kinsella Jan 28 '12 at 7:35