Complex Integration $ \int \limits_{-\infty}^{+\infty} \! e^{-jx\nu} {\lambda^2 \over (\lambda^2 + \nu^2)} \, \mathrm d\nu $ I have the following integration to do and i can't find a way to do it, i tried integration by parts but that didn't help.
$$ \int \limits_{-\infty}^{+\infty} \! e^{-jx\nu} {\lambda^2 \over (\lambda^2 + \nu^2)} \, \mathrm d\nu $$
Any help is appreciated.

You consider the following integral in the complex plane:
$$ \lambda^2 \oint_C dz\:\frac{e^{-i x z}}{\lambda^2 + z^2} $$
Note that I am assuming you meant by $j$ what we mean by $i$.  For $x>0$ we take as $C$ the closed semicircle in the lower half plane.  In this case, the integral about the semicircular arc goes to zero as the radius goes to $\infty$.  The integral is $i 2 \pi$ times the sum of the residues of the poles inside $C$, which in this case includes only the pole $z=-i \lambda$.  Therefore
$$\int_{-\infty}^{\infty} dx \: e^{-i x \nu} \frac{\lambda^2}{\lambda^2+\nu^2} = i 2\pi (i \lambda^2/(2 \lambda)) e^{-\lambda x}= -\pi \lambda e^{-\lambda x} \; \; ,x>0$$
For $x<0$ we take as $C$ the semicircle in the upper half plane; the analysis is the same.  The result for any value of $x$ is
$$\int_{-\infty}^{\infty} dx \: e^{-i x \nu} \frac{\lambda^2}{\lambda^2+\nu^2} = -\pi \lambda e^{-\lambda |x|} \mathrm{sgn}(x)$$
