# Fourier Analysis – Fourier Transform of e^(-|x|)/?|x|

fourier analysisfourier transformintegration

Please help me find Fourier Transform for $$f(x) = \frac{e^{-|x|}}{\sqrt{|x|}}$$

What I have now:
$$\frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}}{f(x)e^{-ixy}~dx} = \sqrt{\frac{2}{\pi}}\int_{0}^{\infty}{\frac{e^{-x}\cos(xy)}{\sqrt{x}}}~dx = \sqrt{\frac{2}{\pi}}\operatorname{Re}\left(\int_{0}^{\infty}{\frac{e^{ixy-x}}{\sqrt{x}}}~dx\right)$$

I'm stuck on this and do not know what to do next, any help would be greatly appreciated!

We begin by folding the integral and enforcing the substitution $$x\mapsto x^2$$. This reveals that

\begin{align} \int_{-\infty}^\infty \frac{e^{-|x|}}{\sqrt{|x|}}e^{-ixy}\,dx&=2\int_0^\infty \frac{e^{-x}}{\sqrt x}\,\cos(xy)\,dx\\\\ &=2\text{Re}\left(\int_0^\infty \frac{e^{-(1-iy)x}}{\sqrt x}\,dx\right)\\\\ &=4\text{Re}\left(\int_0^\infty e^{-\left(\sqrt{1-iy}\,x\right)^2}\,dx\right)\\\\ \end{align}

Next, we enforce the substitution $$\sqrt{1-iy}\,x\mapsto x$$ to obtain

$$\int_{-\infty}^\infty \frac{e^{-|x|}}{\sqrt{|x|}}e^{-ixy}\,dx=4\text{Re}\left(\frac1{\sqrt{1-iy}}\int_0^{\infty/\sqrt{1-iy}} e^{-x^2}\,dx\right)$$

Invoking Cauchy's Integral Theorem, we can deform the contour back onto the real line to find

\begin{align} \int_{-\infty}^\infty \frac{e^{-|x|}}{\sqrt{|x|}}e^{-ixy}\,dx&=4\text{Re}\left(\frac1{\sqrt{1-iy}}\int_0^{\infty} e^{-x^2}\,dx\right)\\\\ &=2\sqrt{\pi}\text{Re}\left(\frac1{\sqrt{1-iy}}\right)\\\\ &=\frac{2\sqrt{\pi}}{\sqrt{1+y^2}}\text{Re}\left(\sqrt{1+iy}\right)\\\\ &=\frac{2\sqrt{\pi}}{\sqrt{1+y^2}}\text{Re}\left(\sqrt{\sqrt{1+y^2}e^{i\arctan(y)}}\right)\\\\ &=\frac{2\sqrt{\pi}}{\sqrt{1+y^2}} \sqrt{\sqrt{1+y^2}}\cos(\arctan(y)/2)\\\\ &=\frac{2\sqrt{\pi}}{\sqrt{1+y^2}} \sqrt{\sqrt{1+y^2}}\sqrt{\frac{1+\cos(\arctan(y))}{2}}\\\\ &=\sqrt{2\pi}\frac{\sqrt{\sqrt{1+y^2}\left(1+\frac1{\sqrt{1+y^2}}\right)}}{\sqrt{1+y^2}}\\\\ &=\sqrt{2\pi}\sqrt{\frac{\sqrt{1+y^2}+1}{1+y^2}} \end{align}