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.

  • $\begingroup$ 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

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.

  • 2
    $\begingroup$ That's the integral representation I was thinking of when I read the question. +1 $\endgroup$ – Random Variable Sep 9 '17 at 0:09
  • 1
    $\begingroup$ nice $\left(+1\right)$ $\endgroup$ – tired Sep 10 '17 at 15:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.