Inverse Fourier transform of $\frac{1}{k^2+a^2}$

Asked Jul 29 '17 at 00:20

Active Jul 29 '17 at 00:20

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I am trying to take the inverse Fourier transform of $\frac{1}{k^2+a^2}$. I know what the answer is (something like $e^{-|x|}$), but I'm having trouble actually doing the integral in the inverse Fourier transform.

First I noted that $\frac{1}{k^2+a^2}$ is even in $k$, so it's inverse Fourier transform is $$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \frac{1}{k^2+a^2} \cos(kx) dk$$

Where do I go from here? I tried integration by parts, but that didn't lead anywhere. And looking at the Fourier transform of $e^{-|x|}$ didn't really help.

Ideas?

asked Jul 29 '17 at 00:20

Kurt

1,120

In order to write a pertinent answer, do you know about Contour Integration? – Philip Jul 29 '17 at 01:14
Unfortunately no, I don't. – Kurt Jul 29 '17 at 01:20
$\cos (k x)=\frac{1}{2} e^{-i k x}+\frac{1}{2} e^{i k x}\quad$ maybe this helps – Raffaele Jul 29 '17 at 16:23

Inverse Fourier transform of $\frac{1}{k^2+a^2}$

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