1. ## Residue theorem

Hi I'm having some problems with the below question

Use the residue theorem to compute

$\int\limits_{-\infty}^{\infty} \frac{\cos(x)}{1+x^2} dx$

So i rewrote the intergral as

$\int\limits_{-\infty}^{\infty} \frac{\cos(x)}{1+x^2} dx = \Re(\int\limits_{-\infty}^{\infty} \frac{e^{jx}}{1+x^2} dx)$

Where do i go from here?

Thanks

2. The value of the complex integral...

$\int_{-\infty}^{+\infty} \frac{e^{j x}}{1+x^{2}}\cdot dx$ (1)

... is given by...

$\lim_{R \rightarrow \infty} \int_{\Gamma} f(z)\cdot dz$ (2)

... where $f(z) = \frac{e^{j z}}{1+z^{2}}$ and $\Gamma$ is the red path represented in figure...

All that is true only if the socalled 'Jordan's lemma' is verified, just as in this case. For the 'second Cauchy integral formula' is...

$\int_{\Gamma} f(z)\cdot dz = 2 \pi j \sum_{i} r_{i}$ (3)

... where $r_{i}$ is the residue relative at each pole $z_{i}$ of $f(*)$ inside $\Gamma$ and it is given by...

$r_{i} = \lim_{z \rightarrow z_{i}} (z-z_{i})\cdot f(z)$ (4)

In our case the only pole of $f(*)$ inside $\Gamma$ is $z= + j$ so that is...

$r_{1} = \lim_{ z \rightarrow + j} (z - j)\cdot \frac{e^{j z}}{1+z^{2}} = \frac{e^{-1}}{2 j}$ (5)

... and finally...

$\int_{- \infty}^{+ \infty} \frac {\cos x}{1+x^{2}}\cdot dx = 2 \pi j \cdot r_{1} = \frac{\pi}{e}$ (6)

Kind regards

$\chi$ $\sigma$