# Integral $\int_{-\infty}^\infty \frac{e^{ikx}}{x^{3/2}}dx$

Hi I'm trying to solve this integral Fourier Transform $$\int_{-\infty}^\infty \frac{e^{ikx}}{x^{3/2}}dx=\sqrt{2\pi|k|}(1+i) (-1+\text{sgn}(k))$$ where sgn(k)$=1$ for k>1 and $-1$ for k<1. I am trying to use residues. Thanks there is a singularity at $x=0$. We can try and write $$e^{ikx}=\cos x+i\sin x$$ but I don't think it will help. IT will be nice to use a contour with $e^{ikx}$ instead. Thanks

• I think you'll be using the famous keyhole contour. Break in the integral up into $(0,\infty)$ and $(-infty,0)$. You should be able to write the latter in terms of the former. To actually compute the former, draw a keyhole (sometimes called a pacman) contour and be mindful of the branch cut. Without actually working it out, I believe this is the typical crank to turn for this type of integral. – Jason Mar 12 '14 at 6:58

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With $\ds{s \equiv \ic x\quad\imp\quad x = -\ic s = \expo{3\ic\pi/2}s}$: \begin{align} \int_{-\infty + \ic 0^{+}}^{\infty + \ic 0^{+}} {\expo{\ic kx} \over x^{3/2}}\,\dd x&= \Theta\pars{-k}\int_{-0^{+} - \ic\infty}^{-0^{+} + \ic\infty} {\expo{\ic k\pars{-\ic s}} \over \pars{\expo{3\ic\pi/2}s}^{3/2}}\,\pars{-\ic\,\dd s} \\[3mm]&= \Theta\pars{-k}\pars{-\ic\expo{-9\ic\pi/4}} \int_{0^{+} -\ic\infty}^{0^{+} + \ic\infty}{\expo{\verts{k}s} \over \expo{-3\ic\pi/2}s^{3/2}}\,\dd s \\[3mm]&=\Theta\pars{-k}{\root{2} \over 2}\pars{-1 + i} \int_{0^{+} -\ic\infty}^{0^{+} + \ic\infty}s^{-3/2}\expo{\verts{k}s}\,\dd s\tag{1} \end{align}
$$\int_{0^{+} - \ic\infty}^{0^{+} + \ic\infty}s^{-3/2}\expo{\verts{k}s}\,\dd s =\lim_{\epsilon \to 0^{+}}\pars{% {\cal I}_{+} + {\cal I}_{-}}\tag{2}$$
We replace $\pars{3}$ and $\pars{4}$ in $\pars{2}$. In the limit $\epsilon \to 0^{+}$: \begin{align} \Theta\pars{-k}\int_{0^{+} - \ic\infty}^{0^{+} + \ic\infty}s^{3/2}\expo{\verts{k}s}\,\dd s &=\Theta\pars{-k}\bracks{4\ic\verts{k}\int_{0}^{\infty}x^{-1/2}\expo{-\verts{k}x}\,\dd x} \\[3mm]&=4\ic\root{\verts{k}}\Theta\pars{-k}\ \underbrace{\int_{0}^{\infty}x^{-1/2}\expo{-x}\,\dd x} _{\ds{\Gamma\pars{1/2} = \root{\pi}}} \\[3mm]&=\Theta\pars{-k}\bracks{4\ic\root{\pi\verts{k}}} =-2\ic\bracks{-1 + \sgn\pars{k}}\,\root{\pi\verts{k}}\tag{5} \end{align}
We'll replace $\pars{5}$ in $\pars{1}$: \begin{align} \color{#00f}{\large\int_{-\infty}^{\infty}{\expo{\ic kx} \over x^{3/2}}\,\dd x}&= {\root{2} \over 2}\pars{-1 + i}\braces{% -2\ic\bracks{-1 + \sgn\pars{k}}\,\root{\pi\verts{k}}} \\[3mm]&=\color{#00f}{\large\root{2\pi\verts{k}}\pars{1 + i}\bracks{-1 + \sgn\pars{k}}} \end{align}