$$\int_0^{\infty} \frac{\cos(kx)}{k^2 + a^2} dk$$
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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1"using taylor expansion" - nope; you'll be getting not a few divergent integrals that way. – J. M. ain't a mathematician Apr 03 '13 at 17:09
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@Aryabhata: no it isn't. It's standard for an inverse Fourier transform. – Ron Gordon Apr 03 '13 at 17:31
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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.
Ron Gordon
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Excuse-me @Ron Gordon, but could you elaborate a little on which integral of the complex plane should be integrated ? Exactly the same, but with $k$ a variable of $\mathbb{C}$? – Rhaena Apr 26 '21 at 14:28
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@Rhaena: yes, and the contour can be a semicircle in the upper half plane. – Ron Gordon Apr 26 '21 at 14:29
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Mmh so $C_R := {z \in \mathbb{C} | |z| = R, \Im(z) > 0}$ can be usefull here ? – Rhaena Apr 26 '21 at 14:31