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

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, TMM Apr 3 '13 at 17:46

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"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
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1 Answer 1

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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