# Evaluating $\int_0^{\infty} \frac{\cos(kx)}{k^2 + a^2} dk$ [duplicate]

$$\int_0^{\infty} \frac{\cos(kx)}{k^2 + a^2} dk$$

This equals to $\displaystyle\frac{1}{2} \int_{-\infty}^{\infty} \frac{\cos(kx)}{k^2 + a^2} dk$ and I solved it, but the answer is not of exponential form. How do I evaluate this in exponential form?

-

## marked as duplicate by Aryabhata, Marvis, muzzlator, Davide Giraudo, TMMApr 3 '13 at 17:46

$$\int_{-\infty}^{\infty} dk \: \frac{e^{i k x}}{k^2+a^2} = \frac{\pi}{a} e^{-a |x|}$$