I am trying to do

$$\int_{-\infty}^\infty e^{iax} (x-c)^{-p/q} dx$$

for co-prime $p,q>0$ and $c\in \mathbb{C}$, $\mathrm{Re}(c)=0$. The case $q=1$ is in the complex analysis textbooks ($c$ is a pole of order $p$ then). $q=2$ and $p=1$ is slightly harder, but still not bad: a branch cut along the $i$-axis, two large quarter-circles, a small circle around $c$, and Jordan Lemma do the job (look here for even more general 2 roots in the denominator). I could not find anything on the general case, even for $q=2$ and $p>2$.

The case $q=1$ uses

$$\oint_C \frac{e^{iax}}{(x-c)^p} dx=\frac{2\pi i}{(p-1)!} \left(\frac{d^{p-1}}{d x^{p-1}} e^{iax}\right)\bigg|_{x=c}$$

for a contour $C$ around $c$. This brought me to the somewhat esoteric "Fractional calculus" which more or less takes the above as the definition of a generalized derivative. But I am not sure if there is anything like "Complex fractional calculus" and how does that help me with the integral. At this point, I have the feeling that my lack of basic understanding of complex analysis might have taken me too far afield...

It seems to me that $\Gamma$ function, Riemann surface etc will have to appear in the solution, but I am not sure how.


Your answer heads in the right direction, but perhaps working harder than turns out to be necessary. There is a by-now-standard computation of the Fourier transform of $(x-w)^\alpha$ for $w\not\in \mathbb R$ for any $\alpha\in\mathbb R$, at first for $\Re(\alpha)<-1$ for literal integrability, and then for arbitrary real $\alpha$, by meromorphic continuation, etc.

The computation in the range of convergence invokes a widely useful trick involving $\Gamma(s)$: for $\Re(s)>0$ and $y>0$, $\int_0^\infty t^s\,e^{-ty}\;dt/t= y^{-s}\Gamma(s)$, by changing variables. By the Identity Principle from Complex Analysis, the same holds for $y$ complex with $\Re(y)>0$. Thus, replacing $y$ by $y-2\pi i x$, $$ \int_0^\infty t^{s-1}\,e^{-ty}\cdot e^{-2\pi ixt}\;dt \;=\; (y-2\pi ix)^{-s}\,\Gamma(s) $$ That is, letting $f_y(t)$ be the function that is $t^{s-1}e^{-ty}$ for $t>0$, and $0$ for $t<0$, the Fourier transform of $f_y$ at $x$ is $(y-2\pi ix)^{-s}\,\Gamma(s)$.

Then Fourier inversion gives the sort of result you want...

  • $\begingroup$ Great! I felt I was working too hard and that there should be a neater way to do this. Anyway, got to learn some complex analysis. $\endgroup$ – user3763801 Mar 2 '18 at 5:29

Here is what I worked up for $q=2$. For $x>0$

$ \DeclareMathOperator{\Res}{Res} $

\begin{gather*} f(x)&\stackrel{def}=\frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{e^{itx}}{(1+it)^{\frac{n}{2}}}dt=\frac{ x^{\frac{n}{2}-1} e^{-x}}{\Gamma\left(\frac{n}{2}\right)} \end{gather*}

This is not surprising, because I was actually trying to get the $\chi^2_n$-distribution by inverting its characteristic function. In this case the denominator looks like $1+2it$ which results in a $2^{-\frac{n}{2}}$ factor in the result.


\begin{align*} f(x)\stackrel{i(\frac{z}{x}+1)}{=} \frac{i x^{\frac{n}{2}-1}e^{-x}}{2 \pi (-1)^\frac{n}{2}}\int_{-x+i\infty}^{-x-i\infty} z^{-\frac{n}{2}}e^{-z}dz \end{align*}

I needed the following two.

More or less Jordans' Lemma: \begin{align*} \int_{-\pi/2}^{\pi/2} e^{-x \cos\theta}d\theta&=\int_0^{\pi} e^{-x \sin\theta}d\theta= 2\int_0^{\pi/2} e^{-x\sin\theta}d\theta\leq 2\int_0^{\pi/2} e^{- 2 x\theta/\pi}d\theta\\ &\stackrel{-\pi\theta/{2x}}{=}\frac{\pi}{2 x}\int_{-x}^0 e^{\theta}d\theta=O(x^{-1}) \end{align*}

This is kind of trivial: \begin{align*} \int_{0}^{2 \pi} e^{i a \theta}e^{-b e^{i \theta}}d\theta &=\frac{1}{ia} \int_{0}^{2 \pi} e^{-b e^{i \theta}}d e^{i a \theta} =-\frac{1}{ia} \int_{0}^{2 \pi} e^{i a\theta }d e^{-b e^{i \theta}}\\ &=\frac{b}{a} \int_{0}^{2\pi} e^{i (a+1) \theta} e^{-b e^{i \theta}} d\theta \end{align*}

\begin{gather*} g(z)\stackrel{def}{=}z^{-\frac{n}{2}}e^{-z} \end{gather*}

Case $n=2m$.

\begin{gather*} C_1=[R, -R]i-x,\ C_2=Re^{i [-\pi/2, \pi/2]}-x \end{gather*}

\begin{align*} \Res(g(z), 0)=\frac{1}{\Gamma(m)}\frac{ d^{m-1} }{dz^{m-1}} e^{-z} \bigg|_{0}=\frac{(-1)^{m-1}}{\Gamma(m)} =\frac{1}{2\pi i} \left( \int_{C_1} g(z) dz + \int_{C_2} g(z) dz \right) \end{align*}

\begin{align*} \left|\int_{C_2} g(z) dz\right|&\leq \frac{R e^x}{(R-x)^m} \int_{-\pi/2}^{\pi/2} e^{-R \cos\theta} d\theta=O(R^{-m}) \end{align*}

\begin{gather*} f(x)= \frac{i x^{\frac{n}{2}-1} e^{-x}}{2 \pi (-1)^m} \cdot 2\pi i\cdot \frac{(-1)^{m-1}}{\Gamma(m)} = \frac{x^{\frac{n}{2}-1}e^{-x}}{\Gamma(\frac{n}{2})} \end{gather*}

Case $n=2m+1$.

\begin{gather*} C_1=[R, -R]i-x, C_2=Re^{i [-\pi/2, 0]}-x, C_3=[R-x,\varepsilon]\\ C_4=\varepsilon e^{i[2\pi, 0]}, C_5=[\varepsilon, R-x], C_6=Re^{i [0, \pi/2]}-x \end{gather*}

\begin{align*} \left|\int_{C_2} g(z) dz\right|&\leq \frac{R e^x}{(R-x)^\frac{n}{2}}\int_{-\pi/2}^0 e^{-R\cos\theta}d\theta=O(R^{-m}) \end{align*}

Similar for $C_6$.

\begin{align*} \left|\int_{C_4} g(z) dz\right| &=\varepsilon^{1-n/2} \left|\int_{2\pi}^{0} e^{i\theta}e^{-\varepsilon e^{i\theta}} e^{-n i\theta/2}d\theta\right| =\varepsilon^{1/2-m} \left|\int_0^{2\pi} e^{i(1/2-m)\theta}e^{-\varepsilon e^{i\theta}} d\theta\right|\\ &=\frac{\varepsilon^{1/2}}{(-1/2)_m} \left|\int_{0}^{2\pi} e^{i\theta/2}e^{-\varepsilon e^{i\theta}} d\theta\right| =O(\varepsilon^{1/2}) \end{align*}

\begin{align*} \lim \int_{C_5} g(z) dz&=\Gamma\left(1-\frac{n}{2}\right)=\frac{\pi }{\Gamma\left(\frac{n}{2}\right) \sin {\frac{\pi n}{2}}} =\frac{(-1)^{\frac{n-1}{2}} \pi }{\Gamma\left(\frac{n}{2}\right)} =\frac{ \pi (-1)^{\frac{n}{2}}}{i \Gamma\left(\frac{n}{2}\right)} \end{align*}

\begin{align*} \lim \int_{C_3} g(z) dz&= \int_\infty^0 (-1)^n t^{-\frac{n}{2}}e^{-t} dt = (-1)^{n-1} \int_0^\infty t^{-\frac{n}{2}}e^{-t} dt=\lim \int_{C_5} g(z) dz \end{align*}

\begin{gather*} f(x)= \frac{i x^{\frac{n}{2}-1} e^{-x}}{2 \pi (-1)^{\frac{n}{2}}}\cdot 2\cdot \frac{ \pi (-1)^{\frac{n}{2}} }{i \Gamma\left(\frac{n}{2}\right)} =\frac{ x^{\frac{n}{2}-1} e^{-x}}{\Gamma\left(\frac{n}{2}\right)} \end{gather*}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.