Fourier transform of a Lévy density $\frac{1}{\sqrt{2\pi }}\int_{0}^{\infty} e^{ikx-\frac{1}{2x}}x^{-\frac{3}{2}}dx$

Question

A Lévy density is defined as

$$q(x;1/2,1)=\frac{1}{\sqrt{2\pi }}e^{-\frac{1}{2x}}x^{-\frac{3}{2}}$$

for $x>0$

I am looking for it's Fourier transform:

$$g(k;1/2,1)=\frac{1}{\sqrt{2\pi }}\int_{0}^{\infty} e^{ikx-\frac{1}{2x}}x^{-\frac{3}{2}}dx = e^{-\sqrt{|k|}(1-i\text{ sign}(k))} $$

where sign$(k) = k/|k|$

How to evaluate this integral?

The author from the book I took it from, suggests splitting the integral into real and imaginary part (V.V. Uchaikin, V.M. Zolotarev, Chance and Stability. Stable Distributions and their Applications). Unfortunatly, he ommits the evaluation, giving only the final result.

Answer 1

Let $I(k)$ be the integral given by

$$\begin{align} I(k)&=\int_0^\infty e^{ikx-1/2x} x^{-3/2}\,dx\\\\ \end{align}$$

We assume that $k>0$ and leave the case for which $k<0$ to the reader.

Let $a=(1-i)\sqrt k$. Then, we can write $I(k)$ as

$$I(k)=\int_0^\infty e^{-\frac a2 \left(ax+\frac1{ax}\right)}\,x^{-3/2}\,dx$$

Enforcing the substitution $ax\to x$ yields

$$\begin{align} I(k)&=\sqrt a\int_0^\infty e^{-\frac a2 \left(x+\frac1{x}\right)}\,x^{-3/2}\,dx\\\\ &=\sqrt a\,e^{-a}\int_0^\infty e^{-\frac a2 \left(\sqrt x-\frac1{\sqrt x}\right)^2}\,x^{-3/2}\,dx \tag 1 \end{align}$$

Now, if we enforce the substitution $x\to 1/x$, we find that

$$I(k)=\sqrt a\,e^{-a}\int_0^\infty e^{-\frac a2 \left(\sqrt x-\frac1{\sqrt x}\right)^2}\,x^{-1/2}\,dx \tag 2$$

Adding $(1)$ and $(2)$ and dividing by $2$ yields

$$\begin{align} I(k)&=\sqrt a\,e^{-a}\int_0^\infty e^{-\frac a2 \left(\sqrt x-\frac1{\sqrt x}\right)^2}\,\left(\frac12 x^{-1/2}+\frac12 x^{-3/2}\right)\,dx\\\\ &=\sqrt a\,e^{-a}\int_0^\infty e^{-\frac a2 \left(\sqrt x-\frac1{\sqrt x}\right)^2}\,\frac{d}{dx}\left(\sqrt x-\frac1{\sqrt x}\right)\,dx\\\\ &=\sqrt a\,e^{-a}\int_{-\infty}^\infty e^{-\frac a2 x^2}\,dx\\\\ &=\sqrt{2\pi}e^{-a}\\\\ &=\sqrt{2\pi}e^{(1-i)\sqrt k} \end{align}$$

Answer 2

This is a sort of heuristic approach, but it seems more straightforward to work with Laplace transforms and then extend the result to the given Fourier integral.

I reproduce the solution of the problem of finding the inverse Laplace transform of $e^{-\sqrt{s}}$, i.e.,

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, e^{-\sqrt{s}} e^{z t} $$

given here below.

You can use a contour integration by deforming the Bromwich contour about the negative real axis and exploiting a branch cut of $\sqrt{z}$ about that axis. So, consider the integral

$$\oint_C dz \: e^{-\sqrt{z}} e^{z t}$$

where $C$ is a keyhole contour about the negative real axis, as pictured below.

We will define $\text{Arg}{z} \in (-\pi,\pi]$, so the branch is the negative real axis. There are $6$ pieces to this contour, $C_k$, $k > \in \{1,2,3,4,5,6\}$, as follows.

$C_1$ is the contour along the line $z \in [c-i R,c+i R]$ for some large value of $R$.

$C_2$ is the contour along a circular arc of radius $R$ from the top of $C_1$ to just above the negative real axis.

$C_3$ is the contour along a line just above the negative real axis between $[-R, -\epsilon]$ for some small $\epsilon$.

$C_4$ is the contour along a circular arc of radius $\epsilon$ about the origin.

$C_5$ is the contour along a line just below the negative real axis between $[-\epsilon,-R]$.

$C_6$ is the contour along the circular arc of radius $R$ from just below the negative real axis to the bottom of $C_1$.

We will show that the integral along $C_2$,$C_4$, and $C_6$ vanish in the limits of $R \rightarrow \infty$ and $\epsilon \rightarrow 0$.

On $C_2$, the real part of the argument of the exponential is

$$R t \cos{\theta} - \sqrt{R} \cos{\frac{\theta}{2}}$$

where $\theta \in [\pi/2,\pi)$. Clearly, $\cos{\theta} \lt 0$ and $\cos{\frac{\theta}{2}} \gt 0$, so that the integrand exponentially decays as $R \rightarrow \infty$ and therefore the integral vanishes along $C_2$.

On $C_6$, we have the same thing, but now $\theta \in (-\pi,-\pi/2]$. This means that, due to the evenness of cosine, the integrand exponentially decays again as $R \rightarrow \infty$ and therefore the integral also vanishes along $C_6$.

On $C_4$, the integral vanishes as $\epsilon$ in the limit $\epsilon \to 0$. Thus, we are left with the following by Cauchy's integral theorem (i.e., no poles inside $C$):

$$\left [ \int_{C_1} + \int_{C_3} + \int_{C_5}\right] dz \: e^{-\sqrt{z}} e^{z t} = 0$$

On $C_3$, we parametrize by $z=e^{i \pi} x$ and the integral along $C_3$ becomes

$$\int_{C_3} dz \: e^{-\sqrt{z}} e^{z t} = e^{i \pi} \int_{\infty}^0 dx \: e^{-i \sqrt{x}} e^{-x t}$$

On $C_5$, however, we parametrize by $z=e^{-i \pi} x$ and the integral along $C_5$ becomes

$$\int_{C_5} dz \: e^{-\sqrt{z}} e^{z t} = e^{-i \pi} \int_0^{\infty} dx \: e^{i \sqrt{x}} e^{-x t}$$

We may now write

$$-\frac{1}{i 2 \pi} \int_0^{\infty} dx \: e^{- x t} \left ( e^{i \sqrt{x}} - e^{-i \sqrt{x}} \right ) + \frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: e^{-\sqrt{s}} e^{s t} = 0$$

Therefore, the ILT of $\hat{f}(s) = e^{-\sqrt{s}}$ is given by

$$\begin{align}\frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: e^{-\sqrt{s}} e^{s t} &= \frac{1}{i 2 \pi} \int_0^{\infty} dx \: e^{- x t} \left ( e^{i \sqrt{x}} - e^{-i \sqrt{x}} \right )\\ &= \frac{1}{\pi} \int_{-\infty}^{\infty} du\: u \,e^{-t u^2} \sin{u}\end{align}$$

The last step involved substituting $x=u^2$ and exploiting the evenness of the integrand. This integral may be evaluated as follows:

$$\begin{align}\frac{1}{\pi} \int_{-\infty}^{\infty} du\: u \,e^{-t u^2} \sin{u} &= \frac{1}{\pi} \Im{\left [\int_{-\infty}^{\infty} du\:u\, e^{-t u^2} e^{i u} \right]}\\ &= \frac{1}{\pi} \Im{\left [\int_{-\infty}^{\infty} du\:u\, e^{-t (u-i/(2 t))^2} e^{-1/(4 t)}\right ]}\\ &= \frac{1}{\pi} e^{-1/(4 t)} \Im{\left [\int_{-\infty}^{\infty} dv \: \left ( v + \frac{i}{2 t} \right ) e^{-t v^2} \right]}\\ &= \frac{1}{\pi} e^{-1/(4 t)} \frac{1}{2 t} \sqrt{\frac{\pi}{t}} \end{align}$$

Therefore the result is that

$$\mathcal{L}^{-1}[e^{-\sqrt{s}}](t) = \frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: e^{-\sqrt{s}} e^{s t} = \frac{1}{2 \sqrt{\pi}} t^{-3/2} e^{-\frac{1}{4 t}}$$

So, first we see that

$$\frac1{2 \sqrt{\pi}} \int_0^{\infty} dx \, e^{-s x} e^{-\frac1{4 x}} x^{-3/2} = e^{-\sqrt{s}}$$

We may then observe that if we sub $x \mapsto x/2$ and $s \mapsto 2 s$, we get

$$\frac1{\sqrt{2 \pi}} \int_0^{\infty} dx \, e^{-s x} e^{-\frac1{2 x}} x^{-3/2} = e^{-\sqrt{2 s}}$$

Now consider $s=-i k = e^{i 3 \pi/2} k$ for $k \gt 0$ and $s=+i k = e^{i \pi/2} k$ for $k \lt 0$. Then the stated result is reproduced.