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How to evaluate this integral?

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

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

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    http://en.wikipedia.org/wiki/Residue_theorem#Example2012-04-21
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    Replace $\cos x$ by $\operatorname{Re} e^{ix}$.2012-04-21
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    yes, this integral is with the Cauchy's method of residues. I don't remenber. Long time, i did not see this.2012-04-21
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    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

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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?

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    thanks for reminding me, I'll finish the details, I need to see the way2012-04-21
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    Sure, if you need any more help, don't hesitate to ask.2012-04-21