1
$\begingroup$

Possible Duplicate:
How to evaluate this integral?

Let $b>0$. Evaluate the next value of integral:

$\begin{equation} \int_{-\infty} ^{\infty} \dfrac{\cos x}{x^2 + b^2} \ dx \end{equation}$

  • 0
    I remembering doing this same exact integral in class. Wierd. The answer comes out as $\displaystyle\frac{\pi}{be^{b}}$ after you evaluate the residue at $x = ib$2012-04-22

1 Answers 1

2

If you're familiar with the residue theorem, you know $ \int_{-\infty}^\infty R(x)e^{ix}dx=2\pi i\sum_{y>0}\mathrm{Res}R(z)e^{iz} $ where $R(x)$ is some rational function. However, the real part of this integral is $ \int_{-\infty}^\infty R(x)\cos(x)dx, $ so the integral you want is just the real part of $ 2\pi i\sum_{y>0}\mathrm{Res}\frac{e^{iz}}{z^2+b^2}. $ Note that the sum is over residues in the upper half plane. Note that the poles occur at $\pm bi$, and since $b>0$, the only residue you need to worry about is at $bi$. Can you proceed?

  • 0
    Sure, if you need any more help, don't hesitate to ask.2012-04-21