
Doubly Infinite Integral
Show that for any real numbers $\displaystyle a, b, c$ with $\displaystyle a < b < c$ we have
$\displaystyle PV \int^{\infty}_{\infty} \frac{1}{(xa)(xb)(xc)}dx =0.$
I know, in general that the Cauchy principal value of a doubly infinite integral of a function $\displaystyle f$ is
$\displaystyle PV \int_{\infty}^{\infty}f(x)dx=\lim_{R \rightarrow \infty}\int_{R}^Rf(x)dx$.
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.

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 halfplane. We then have:
$\displaystyle \oint f(z)dz=P.V. \int_{\infty}^{\infty} f(z)dz\pi i(r_1+r_2+r_3)=0$
(since we're only going around $\displaystyle \pi$ at each pole and in the reverse direction). Then:
$\displaystyle P.V.\int_{\infty}^{\infty} \frac{1}{(xa)(xb)(xc)}dx=\pi i (r_1+r_2+r_3)$
where $\displaystyle r_1,r_2,r_3$ 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 semicircle of the contour tends to zero as $\displaystyle R\to\infty$.