Fourier transform of function

Question

What is Fourier transform of $$f(x)=\frac{e^{-|x|}}{\sqrt{|x|}}?$$ I tried to calculate it using $$F(e^{-|x|})=\sqrt{\frac{\pi}{2}}e^{-|a|}$$ and $$F(\frac{1}{\sqrt{|x|}})=\frac{1}{\sqrt{|a|}}$$ and convolution, but it seems to even more complicated.

Answer 1

Your FT is

$$\begin{align}\int_{-\infty}^{\infty} dx \, |x|^{-1/2} \, e^{-|x|} e^{i k x} &= \int_{-\infty}^{0} dx \, (-x)^{-1/2} \, e^{(1+i k) x} + \int_{0}^{\infty} dx \, x^{-1/2} \, e^{-(1-i k) x}\\ &= 2 \int_{0}^{\infty} du \, \left (e^{-(1-i k) u^2} + e^{-(1+i k) u^2}\right )\\ &= \sqrt{\pi} \left [(1-i k)^{-1/2}+(1+i k)^{-1/2} \right ] \\ &= 2 \sqrt{\pi} \, \Re{[(1+i k)^{-1/2}]}\\ &=2 \sqrt{\pi} (1+k^2)^{-1/4} \cos{\left(\frac12 \arctan{k}\right )}\\ &= \sqrt{2 \pi} \frac{\sqrt{1+\sqrt{1+k^2}}}{ \sqrt{1+k^2}}\end{align}$$

Note that I did not rely on the convolution theorem. The lesson here is that sometimes it is easier just to evaluate the FT directly.

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large a}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \fermi\pars{x} & \equiv {\expo{-\verts{x}} \over \root{\verts{x}}} =\int_{-\infty}^{\infty} \tilde{\fermi}\pars{k}\expo{-\ic kx} \,{\dd k \over 2\pi} \\[3mm] & \imp\quad\tilde{\fermi}\pars{k} = \int_{-\infty}^{\infty}{\expo{-\verts{x}} \over \root{\verts{x}}}\,\expo{\ic k x}\,\dd x \end{align}

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\begin{align} \color{#00f}{\large\tilde{\fermi}\pars{k}}&= \int_{-\infty}^{\infty}{\expo{-\verts{x}} \over \root{\verts{x}}}\, \expo{\ic k x}\,\dd x =\int_{-\infty}^{\infty}{\expo{-\verts{x}} \over \root{\verts{x}}}\,\cos\pars{kx} \,\dd x \\[3mm] & = 2\Re\int_{0}^{\infty}{\expo{-x} \over \root{x}}\, \expo{\ic kx}\,\dd x \\[3mm]&=2\Re\int_{0}^{\infty}x^{-1/2}\expo{-\pars{1 - \ic k}x}\,\dd x \\[3mm] & = 2\Re\bracks{\pars{1 - \ic k}^{-1/2}\ \overbrace{\int_{0}^{\infty}x^{-1/2}\expo{-x}\,\dd x} ^{\ds{\Gamma\pars{\half} = \root{\pi}}}} \\[3mm]&=2\root{\pi}\Re\pars{1 - \ic k}^{-1/2} =\bracks{\root{1 + k^{2}}\exp\pars{-\ic\arctan\pars{k}}}^{-1/2} \\[3mm]&=2\root{\pi}\pars{1 + k^{2}}^{-1/4}\cos\pars{\arctan\pars{k} \over 2} \\[3mm] & =2\root{\pi}\pars{1 + k^{2}}^{-1/4} \root{1 + \cos\pars{\arctan\pars{k}} \over 2} \\[3mm]&=\root{2\pi}\pars{1 + k^{2}}^{-1/4} \root{1 + {1 \over \root{\tan^{2}\pars{\arctan\pars{k}} + 1}}} \\[3mm]&=\root{2\pi}\pars{1 + k^{2}}^{-1/4} \root{1 + {1 \over \root{k^{2} + 1}}} \\ [3mm] & =\root{2\pi} \root{{1 \over \root{1 + k^{2}}}\,{1 + \root{1 + k^{2}} \over \root{1 + k^{2}}}} \\[3mm]&=\color{#00f}{\large\root{2\pi}\,\root{1 + \root{1 + k^{2}} \over 1 + k^{2}}} \end{align}

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \tilde{\on{f}}\pars{k} & \equiv \bbox[5px,#ffd]{\int_{-\infty}^{\infty}{% \expo{-\verts{x}} \over \root{\verts{x}}}\,\expo{\ic k x}\,\dd x} = 2\,\Re\int_{0}^{\infty}{% \expo{-x} \over \root{x}}\,\expo{\ic k x}\,\dd x \\[5mm] & = 2\,\Re\int_{0}^{\infty}x^{\color{red}{1/2} - 1}\,\, \expo{-\pars{1 - \ic k}x}\,\,\,\dd x \end{align} Note that $\ds{\expo{-\pars{1 - \ic k}x}\ = \sum_{0}^{\infty}{\bracks{-\pars{1 - \ic k}x}^{\,n} \over n!} = \sum_{0}^{\infty}\color{red}{\pars{1 - \ic k}^{n}} \,\,{\pars{-x}^{\,n} \over n!}}$.

Then, \begin{align} \tilde{\on{f}}\pars{k} & = 2\,\Re\bracks{\Gamma\pars{\color{red}{1 \over 2}} \pars{1 - \ic k}^{-\color{red}{1/2}}}\ \pars{\substack{\ds{Ramanujan's} \\[0.5mm] \ds{Master}\\[0.5mm] \ds{Theorem}}} \\[5mm] & = 2\root{\pi}\Re\braces{\bracks{\root{1 + k^{2}}\expo{\ic\arctan\pars{-k}}\,\,}^{-1/2}} \\[5mm] & = 2\root{\pi}\pars{1 + k^{2}}^{-1/4}\,\, \cos\pars{\arctan\pars{k} \over 2} \\[5mm] & = 2\root{\pi}\pars{1 + k^{2}}^{-1/4}\,\, \root{1 + \cos\pars{\arctan\pars{k}} \over 2} \\[5mm] & = 2\root{\pi}\pars{1 + k^{2}}^{-1/4}\,\, \root{\sec\pars{\arctan\pars{k}} + 1 \over 2\sec\pars{\arctan\pars{k}}} \\[5mm] & = 2\root{\pi}\pars{1 + k^{2}}^{-1/4}\,\, \root{\root{k^{2} + 1} + 1 \over 2\root{k^{2} + 1}} \\[5mm] & = \bbx{\root{2\pi} \root{\root{k^{2} + 1} + 1 \over k^{2} + 1}} \\ & \end{align}