Is there an expression for the inverse Fourier transform of a log-normal function?

Question

i.e. is there a simple solution for the following integral?

$$\int_{-\infty}^{\infty} \exp(-\log^2(|\omega|/\omega_0)) \; \exp(i \omega t) \; d\omega$$

where $\omega_0 > 0$

Failing that, is there a nice series approximation?

Answer 1

Since mention log-normal function, you probably mean to integrate $\omega$ from 0 to $\infty$, instead of using the absolute value.

The answer to your question depends on your acceptable definition of "simple" or, in other words, on the class of functions you would like to represent the integral with. Let's do some rewriting: $$ \mathcal{F}(\omega_0, t) = \omega_0 \mathcal{I}(\omega_0 t) = \omega_0 \int_0^\infty \exp(-\log^2(\omega)) \exp( i \omega \omega_0 t) \mathrm{d} \omega $$ If you insisted on integrating over the real line while using the modulus, notice that $\int_{-\infty}^\infty \exp(-\log^2 \vert \omega/\omega_0 \vert ) \exp(i \omega t) \mathrm{d} \omega = 2 \int_{0}^\infty \exp(-\log^2( \omega/\omega_0 ) ) \cos(\omega t) \mathrm{d} \omega = 2 \omega_0 \mathfrak{Re}\left( \mathcal{I}(\omega_0 t) \right)$.

The integral $\mathcal{I}(t)$ can not be represented in terms of elementary or even hypergeometric functions.

The following formal manipulations give divergent series: $$\begin{eqnarray} \mathcal{I}(t) &=& \int_0^\infty \exp(-\log^2(\omega)) \exp( i \omega t) \mathrm{d} \omega \stackrel{\omega = \exp(u)}{=} \int_{-\infty}^\infty \exp(u-u^2) \exp(i t \mathrm{e}^u ) \mathrm{d} u \\ & \stackrel{\mathrm{formal}}{=}& \sum_{k=0}^\infty \frac{(i t)^k}{k!} \int_{-\infty}^\infty \exp( (k+1) u - u^2) \mathrm{d} u = \sum_{k=0}^\infty \frac{(i t)^k}{k!} \exp\left( \frac{k^2}{4} \right) \end{eqnarray} $$ This formal series, indeed, coincides with the formal Taylor series, since $$ \mathcal{I}^{(k)}(0) = i^k \int_0^\infty \omega^k \exp(-\log^2 \omega) \mathrm{d} \omega \stackrel{\omega = \exp(u)} = i^k \exp\left( \frac{k^2}{4} \right) $$ This means that $\mathcal{I}(t)$ is not analytic at $t=0$.

Gauss-Kronrod quadratures work fine for small $t$, but Levin collocation method might be needed for large $t$. Both methods are implemented in Mathematica:

Answer 2

First of all what you quoted in the question is not really the Fourier transform of a the log-normal distribution. If it was then the function at $t=0$ should give us unity which it does not. But leaving this apart I thought it might be worthwhile to quote another asymptotic expansion of the quantity in question. Denote by $X_\tau$ the the geometric Brownian motion with drift $\mu$ and variance $\sigma^2$ at time $\tau$ subject to $X_0 = x$. Then the following holds true:

\begin{equation} E\left[ e^{\imath k X_\tau} \right| \left. X_0 = x \right] = e^{\imath k x} \sum\limits_{n=0}^\infty \frac{(\imath k)^n}{n!} \cdot {\bar m}_n \end{equation}

where

\begin{equation} {\bar m}_n := \sum\limits_{n_1=0}^n m_{n-n_1} (-x)^{n_1} \binom{n}{n_1} \end{equation}

and $m_n := E[X_\tau^n | X_0=x] = x^n \exp(n(\mu-\sigma^2/2)\tau + n^2 \sigma^2/2 \tau) $ .

Again the series above are divergent but for typical values of drift & volatility truncating the series at the first ten terms gives a good approximation to the result.

{mu, s} = {RandomReal[{0, 1/10}]/256, 
  Sqrt[RandomReal[{0, 1/10}]/256]}; T = 500; t = 50; x = 
 RandomReal[{1, 2}];
k = RandomReal[{0, 1}];
M = 10;
m[n_] := x^n Exp[n (mu - s^2/2) (T - t) + n^2 s^2 (T - t)/2];
bm = Table[
   Sum[m[n - n1] (-x)^n1 Binomial[n, n1], {n1, 0, n}], {n, 0, M}];
Take[Accumulate[
   Exp[I k x] Table[ (I k)^n/n! bm[[1 + n]], {n, 0, 
      M}]], -3] // MatrixForm
NIntegrate[
 Exp[I k z] 1/(s Sqrt[T - t] z) phi[
   1/(s Sqrt[T - t]) (Log[z/x] + (s^2/2 - mu) (T - t))], {z, 0, 
  Infinity}]