Integral of the reciprocal of a quadratic over real line

[wotanub]

Homework Statement

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

Homework Equations

Use contour integration and the residue theorem.

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

Integrate[1/(a*x^2 + b*x + c), {x, -Infinity, Infinity}, Assumptions -> {a > 0, 4*a*c > b^2}]

 

[Ray Vickson]

Homework Statement

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

Homework Equations

Use contour integration and the residue theorem.

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

Integrate[1/(a*x^2 + b*x + c), {x, -Infinity, Infinity}, Assumptions -> {a > 0, 4*a*c > b^2}]

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.

 

[wotanub]

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