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Does anyone have any suggestions for performing the following Fourier transform:

I = $\int_{-\infty}^{\infty} K_0(a \sqrt{x^2+1}) e^{i kx} dx $

where $K_0$ is the modified Bessel function of the second kind.

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  • $\begingroup$ For those interested, I have found the answer in the mean time in the Bateman table of integrals. It is given by, $I(k) = (\frac{\pi}2) (k^2 + a^2)^{-1/2} e^{-(k^2 + a^2)^{1/2}}$. $\endgroup$ – Luca Iliesiu Sep 8 '17 at 22:06
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From the integral representation DLMF $$K_{0}\left(z\right)=\tfrac{1}{2}\int_{0}^{\infty}\exp% \left(-t-\frac{z^{2}}{4t}\right)\frac{\mathrm{d}t}{t}$$, the integral can be written as $$ I(k)=\tfrac{1}{2}\int_{0}^{\infty}\exp\left(-t-\frac{a^{2}}{4t}\right)\frac{dt}{t}\int_{-\infty}^{\infty}\exp\left(ikx-\frac{a^2x^2}{4t}\right)\,dx$$With the Fourier transform of the Gaussian, $$ I(k)=\tfrac{\sqrt{\pi}}{a}\int_{0}^{\infty}\exp\left(-t\left(1+\frac{k^2}{a^2}\right)-\frac{a^{2}}{4t}\right)\frac{dt}{\sqrt{t}} $$ Now, with $u=t\left(1+\frac{k^2}{a^2}\right)$ in this expression, $$I(k)=\tfrac{\sqrt{\pi}}{a} \left(1+\frac{k^2}{a^2}\right)^{-\tfrac{1}{4}}\int_0^\infty \exp\left(-u-\tfrac{k^2+a^2}{4u}\right)\,\frac{du}{\sqrt{u}}$$ Recognizing the integral representation for $K_{-\tfrac{1}{2}}\left(\sqrt{k^2+a^2}\right)$ , the result quoted in the comment follows.

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    $\begingroup$ That's the integral representation I was thinking of when I read the question. +1 $\endgroup$ – Random Variable Sep 9 '17 at 0:09
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    $\begingroup$ nice $\left(+1\right)$ $\endgroup$ – tired Sep 10 '17 at 15:12

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