Integral of the reciprocal of a quadratic over real line

[wotanub]

Homework Statement

This is from Cahill's Physical Mathematics. Exercise 5.23.

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

[itex]\int^{\infty}_{-\infty} \frac{1}{ax^{2}+bx+c} \mathrm{d}x[/itex]

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 [itex]\int^{\infty}_{-\infty} \frac{1}{ax^{2}+bx+c} \mathrm{d}x = \frac{2\pi}{\sqrt{4ac-b^{2}}}[/itex]

My solution:

The roots of the polynomial are

[itex]x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}[/itex]

For convenience, write

[itex]x = \alpha \pm \mathrm{i}\beta[/itex]

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

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

[itex]\int^{\infty}_{-\infty} \frac{1}{ax^{2}+bx+c} \mathrm{d}x[/itex]

[itex]= \oint_{\mathcal{C}} \frac{1}{az^{2}+bz+c} \mathrm{d}z[/itex]

[itex]= \oint_{\mathcal{C}} \frac{1}{(z-(\alpha - \mathrm{i}\beta))(z-(\alpha + \mathrm{i}\beta))} \mathrm{d}z[/itex]

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

[itex] = 2\pi\mathrm{i}\frac{1}{\alpha + \mathrm{i}\beta - (\alpha - \mathrm{i}\beta)}[/itex]

[itex] = 2 \pi\mathrm{i}\frac{1}{2\mathrm{i}\beta}[/itex]

[itex] = \frac{\pi}{\beta}[/itex]

[itex] = \frac{2a\mathrm{i}\pi}{\sqrt{b^{2}-4ac}}[/itex]

[itex] = \frac{2\pi a}{\sqrt{4ac-b^{2}}}[/itex]

So I've got an extra factor of [itex]a[/itex] 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 [itex]a \gt 0[/itex] and [itex]b^{2} – 4ac \lt 0[/itex], use a ghost contour to do the integral

[itex]\int^{\infty}_{-\infty} \frac{1}{ax^{2}+bx+c} \mathrm{d}x[/itex]

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 [itex]\int^{\infty}_{-\infty} \frac{1}{ax^{2}+bx+c} \mathrm{d}x = \frac{2\pi}{\sqrt{4ac-b^{2}}}[/itex]

My solution:

The roots of the polynomial are

[itex]x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}[/itex]

For convenience, write

[itex]x = \alpha \pm \mathrm{i}\beta[/itex]

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

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

[itex]\int^{\infty}_{-\infty} \frac{1}{ax^{2}+bx+c} \mathrm{d}x[/itex]

[itex]= \oint_{\mathcal{C}} \frac{1}{az^{2}+bz+c} \mathrm{d}z[/itex]

[itex]= \oint_{\mathcal{C}} \frac{1}{(z-(\alpha - \mathrm{i}\beta))(z-(\alpha + \mathrm{i}\beta))} \mathrm{d}z[/itex]

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

[itex] = 2\pi\mathrm{i}\frac{1}{\alpha + \mathrm{i}\beta - (\alpha - \mathrm{i}\beta)}[/itex]

[itex] = 2 \pi\mathrm{i}\frac{1}{2\mathrm{i}\beta}[/itex]

[itex] = \frac{\pi}{\beta}[/itex]

[itex] = \frac{2a\mathrm{i}\pi}{\sqrt{b^{2}-4ac}}[/itex]

[itex] = \frac{2\pi a}{\sqrt{4ac-b^{2}}}[/itex]

So I've got an extra factor of [itex]a[/itex] 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.