# Doubly Infinite Integral

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• Dec 14th 2009, 04:42 PM
junipers232
Doubly Infinite Integral
Show that for any real numbers $a, b, c$ with $a < b < c$ we have

$PV \int^{\infty}_{-\infty} \frac{1}{(x-a)(x-b)(x-c)}dx =0.$

I know, in general that the Cauchy principal value of a doubly infinite integral of a function $f$ is

$PV \int_{-\infty}^{\infty}f(x)dx=\lim_{R \rightarrow \infty}\int_{-R}^Rf(x)dx$.

Also, here we have three consecutive real numbers. However, I do not see how we get the Cauchy principal value from all this. I need a bit of help on how this one.
• Dec 15th 2009, 11:38 AM
shawsend
Hi. We can rely on the Residue Theorem to show this by considering a closed contour with indentations around and over each pole along the real axis and then circling back about the upper half-plane. We then have:

$\oint f(z)dz=P.V. \int_{-\infty}^{\infty} f(z)dz-\pi i(r_1+r_2+r_3)=0$

(since we're only going around $\pi$ at each pole and in the reverse direction). Then:

$P.V.\int_{-\infty}^{\infty} \frac{1}{(x-a)(x-b)(x-c)}dx=\pi i (r_1+r_2+r_3)$

where $r_1,r_2,r_3$ are the residues at the poles. I'll leave it to you to show that the sum of those three residues is zero and also should say something about how the integral over the remaining large semi-circle of the contour tends to zero as $R\to\infty$.