According to Mathematica with Fourier transform convention

$$\widehat{f}(\xi)=(2\pi)^{-1/2}\int_{-\infty}^{\infty}f(x)e^{i\pi x}dx$$

The Fourier transform of the function $f(x):=|x|^{-1/2}e^{-|x|}$ is given by


Mathematica spitted out this result very quickly. This is not a function I recognize as having a well-known Fourier transform, but using Fourier analysis, I believe the Fourier transform should be something like (up to a multiplicative constant) the convolution


But Mathematica does not give a nice expression for this integral. I imagine one can evaluate this integral using contour integration and Cauchy residue theorem, but does anyone see a simpler argument, say using Fourier transforms of known functions?


It is easier to do the direct computation.

Forgive me for using my own notation for the time being. Let

$$F(k) = \int_{-\infty}^{\infty} dx \, f(x) e^{i k x}$$


$$f(x) = (2 \pi |x|)^{-1/2} \, e^{-|x|}$$


$$F(k) = \sqrt{\frac{2}{\pi}} \int_0^{\infty} dx \, x^{-1/2} \, e^{-x} \cos{k x} = \sqrt{\frac{8}{\pi}} \int_0^{\infty} dx \, e^{-x^2} \cos{k x^2}$$

The integral is the real part of a complex Gaussian integral:

$$\begin{align}F(k) &= \sqrt{\frac{2}{\pi}} \operatorname{Re}{\left [\int_{-\infty}^{\infty} dx \, e^{-(1-i k) x^2} \right ]} \\ &= \operatorname{Re}{\left (\sqrt{\frac{2}{1-i k}}\right )}\\ &= \sqrt{\frac{2}{\sqrt{1+k^2}}} \cos{\left (\frac12 \arctan{k} \right )}\\ &= \sqrt{\frac{2}{\sqrt{1+k^2}}} \sqrt{\frac{1+\frac1{\sqrt{1+k^2}}}{2}}\end{align}$$


$$F(k) = \sqrt{\frac1{1+k^2}+\frac1{\sqrt{1+k^2}}} $$

as was to be shown.

  • $\begingroup$ I hope you don't mind I posted an answer that is pretty much the same with added details. If you want to add the details to your answer, I will delete my answer. $\endgroup$ – robjohn Jan 29 '16 at 8:50
  • $\begingroup$ @robjohn: I wouldn't do that. Thanks for explaining what you have done. $\endgroup$ – Ron Gordon Jan 29 '16 at 10:03

This answer simply adds some details to Ron Gordon's answer.

Use symmetry to get rid of the absolute values. $$ \begin{align} \int_{-\infty}^\infty\left|x\right|^{-1/2}e^{-\left|x\right|}e^{ixy}\,\mathrm{d}x &=2\operatorname{Re}\left(\int_0^\infty x^{-1/2}e^{-x}e^{ixy}\,\mathrm{d}x\right)\\ &=2\operatorname{Re}\left(\int_0^\infty x^{-1/2}e^{-(1-iy)x}\,\mathrm{d}x\right)\\ \end{align} $$ Since $z^{-1/2}e^{-z}$ has no singularities in the right half-plane, we can use Cauchy's Theorem to get $$ \begin{align} &\int_0^Rx^{-1/2}e^{-(1-iy)x}\,\mathrm{d}x\\ &=\frac1{\sqrt{1-iy}}\int_0^{R(1-iy)}x^{-1/2}e^{-x}\,\mathrm{d}x\\ &=\frac1{\sqrt{1-iy}}\int_0^Rx^{-1/2}e^{-x}\,\mathrm{d}x +\frac1{\sqrt{1-iy}}\int_R^{R(1-iy)}x^{-1/2}e^{-x}\,\mathrm{d}x\\ &\stackrel{R\to\infty}{\to}\frac1{\sqrt{1-iy}}\int_0^\infty x^{-1/2}e^{-x}\,\mathrm{d}x\\ &=\sqrt{\frac\pi{1-iy}} \end{align} $$ since $\Gamma\left(\frac12\right)=\sqrt\pi$ and $$ \int_R^{R(1-iy)}x^{-1/2}e^{-x}\,\mathrm{d}x=O\!\left(R^{1/2}e^{-R}\right) $$ Therefore, $$ \begin{align} \frac1{\sqrt{2\pi}}\int_{-\infty}^\infty\left|x\right|^{-1/2}e^{-\left|x\right|}e^{ixy}\,\mathrm{d}x &=\operatorname{Re}\left(\sqrt{\frac2{1-iy}}\right)\\ &=\bbox[5px,border:2px solid #C0A000]{\sqrt{\frac{1+\sqrt{1+y^2}}{1+y^2}}} \end{align} $$ by solving $(a+ib)^2=\dfrac2{1+y^2}(1+iy)$ for $a$.


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