# Integral of the reciprocal of a quadratic over real line

1. May 31, 2014

### wotanub

1. The problem statement, all variables and given/known data
This is from Cahill's Physical Mathematics. Exercise 5.23.

For $a \gt 0$ and $b^{2} – 4ac \lt 0$, use a ghost contour to do the integral

$\int^{\infty}_{-\infty} \frac{1}{ax^{2}+bx+c} \mathrm{d}x$

2. Relevant equations

Use contour integration and the residue theorem.

3. The attempt at a solution
Mathematica is giving a different result than I got.
It gives $\int^{\infty}_{-\infty} \frac{1}{ax^{2}+bx+c} \mathrm{d}x = \frac{2\pi}{\sqrt{4ac-b^{2}}}$

My solution:

The roots of the polynomial are

$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

For convenience, write

$x = \alpha \pm \mathrm{i}\beta$

where $\alpha \equiv \frac{-b}{2a}, \beta \equiv \frac{-\mathrm{i}}{2a}\sqrt{b^{2}-4ac} \gt 0$

My contour $\mathcal{C}$ will be the real axis, and a large CCW semicircle that encloses the entire UHP. If $z = R \mathrm{e}^{\mathrm{i}\theta}$ then the integral along the arc trivially vanishes as $R→\infty$

$\int^{\infty}_{-\infty} \frac{1}{ax^{2}+bx+c} \mathrm{d}x$

$= \oint_{\mathcal{C}} \frac{1}{az^{2}+bz+c} \mathrm{d}z$

$= \oint_{\mathcal{C}} \frac{1}{(z-(\alpha - \mathrm{i}\beta))(z-(\alpha + \mathrm{i}\beta))} \mathrm{d}z$

$= \mathrm{Res}(\frac{1}{(z-(\alpha - \mathrm{i}\beta))(z-(\alpha + \mathrm{i}\beta))}, z = \alpha + \mathrm{i}\beta)$, since this is the only pole in the UHP

$= 2\pi\mathrm{i}\frac{1}{\alpha + \mathrm{i}\beta - (\alpha - \mathrm{i}\beta)}$

$= 2 \pi\mathrm{i}\frac{1}{2\mathrm{i}\beta}$

$= \frac{\pi}{\beta}$

$= \frac{2a\mathrm{i}\pi}{\sqrt{b^{2}-4ac}}$

$= \frac{2\pi a}{\sqrt{4ac-b^{2}}}$

So I've got an extra factor of $a$ here? Mistakes?
My Mathematica code is
Code (Text):
Integrate[1/(a*x^2 + b*x + c), {x, -Infinity, Infinity}, Assumptions -> {a > 0, 4*a*c > b^2}]

2. May 31, 2014

### Ray Vickson

If $r_1, r_2$ are the roots of the denominator, you need to write $ax^2 + bx + c = a(x-r_1)(x-r_2)$, not just $(x-r_1)(x-r)2)$ as you wrote.

3. May 31, 2014

### wotanub

Oh cool. I know how to contour integrate, I just need to practice multiplying.