# 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?

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

"using taylor expansion" - nope; you'll be getting not a few divergent integrals that way. – J. M. Apr 3 '13 at 17:09
@Aryabhata: no it isn't. It's standard for an inverse Fourier transform. – Ron Gordon Apr 3 '13 at 17:31

You might recognize this as a known Fourier transform:

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

This may be derived via the Residue theorem by considering a similar integral in the complex plane.

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