So, I've been scratching my head over this for the whole day. I'm trying to solve the following integral

$$\int^\infty_{-\infty} \frac{e^{i \alpha (X-\xi)}}{\sqrt{\alpha^2+ \beta^2 }} \, d\alpha$$

$\beta$ here is just a constant. So just for some context, I took a Fourier transformation in order to solve a partial differential equation and after solving and using boundary conditions I obtained an expression in Fourier Space. I am now trying to find the Inverse Fourier transform, which gives the above integral which I can not solve.

I've solved the following using contour integration

$$\int^\infty_{-\infty} e^{i \alpha (X-\xi)} (i \alpha)^{3/4} \, d\alpha$$

Thanks guys.


1 Answer 1


The Fourier transform of $\frac{1}{\sqrt{x^2+\beta^2}}$ is a modified Bessel function of the second kind. For instance,

$$\mathcal{F}\left(\frac{1}{\sqrt{x^2+1}}\right) = \sqrt{\frac{2}{\pi}}\,K_0(|s|).\tag{1}$$

Not by chance, the RHS of $(1)$ is the density function of a normal product distribution.

  • $\begingroup$ Typo? $\frac1{\sqrt{x^2+1}}$ is not an instance of $\frac1{x^2+\beta^2}$. $\endgroup$ Commented Jun 30, 2015 at 17:54
  • $\begingroup$ @DavidC.Ullrich: sure, now fixed. $\endgroup$ Commented Jun 30, 2015 at 17:54
  • $\begingroup$ This would be ok normally. The full integral looks like this. $\endgroup$
    – kisrui
    Commented Jun 30, 2015 at 18:05
  • $\begingroup$ Hmm, this would be fine. However I would still have an exponential left over, which isn't ideal.$p=\frac{\hat{Q}}{\sqrt{\alpha^{2}+\beta^{2}}}$ where, $Q$ is defined to be, $\hat{Q}=\int^{\infty}_{-\infty}A''e^{-i \alpha \xi} d\xi$ Therefore, taking F.T of 1st equation, $\int^{\infty}_{-\infty} \int^{\infty}_{-\infty} A''(\xi) \frac{e^{i \alpha (X-\xi)}}{\sqrt{(\alpha^{2}+ \beta^{2}) }}d\alpha$ Where I will computer A'' numerically, and thus I just need to know how to integrate this: $\int^{\infty}_{-\infty} \frac{e^{i \alpha (X-\xi)}}{\sqrt{(\alpha^{2}+ \beta^{2}) }}d\alpha$ $\endgroup$
    – kisrui
    Commented Jun 30, 2015 at 18:15
  • $\begingroup$ @kisrui: we can deal with the extra exponential through the general property of the Fourier Transform $$\mathcal{F}(f(x)\cdot e^{ki x}) = \mathcal{F}(f)(t+k).$$ $\endgroup$ Commented Jun 30, 2015 at 18:46

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