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.
\frac{1}{a^2+x^2}\ast \frac{1}{a^2+x^2}=\frac{2\pi /a}{4a^2+x^2} (a>0)
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0Thanks 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? ) – 2012-01-28
2 Answers
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 $\hat{E}$?
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0Yes! That is the idea! – 2012-01-28
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.
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0yeah, sorry i meant "use partial fractions." Both have "part" i guess and its been a while since calculus :) – 2012-01-28