# Inverse Fourier transform of $\varphi(t) = \exp\left(\int_0 ^ 1 \frac{e^{itx} - 1}{x} \right)$?

I'm looking for the distribution whose Fourier transform is given by $$\varphi(t) = \exp\left(\int_0 ^ 1 \frac{e^{itx} - 1}{x} \right),$$ where as usual $\varphi(t) = \int_{- \infty} ^ \infty e^{itx} dF(x)$ and $F$ is a probability distribution function on the real line. The distribution should probably be absolutely continuous and put mass on the positive half-line, so we likely want $f$ satisfying $\varphi(t) = \int_0 ^ \infty e^{itx} f(x) \ dx$, if that helps.

I'm not really sure where to begin with this. I don't know that it has an answer in terms of any well known functions. Mathematica choked when trying to invert it and I haven't been able to find any expression for it in any tables that I've looked in.

For some context, this popped up as the characteristic function of an otherwise unknown random variable of interest.

If possible, I would ask that anyone who posts a known solution to include the source.

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I guess the use of inverse Fourier transform formula didn't lead to interesting things. – Davide Giraudo May 3 '12 at 21:20
@Davide do you mean that Mathematica not solving it probably means there isn't a happy solution? It really did try for a long time before giving up. If my friends here don't see anything I may have to try to get at the density numerically. – guy May 3 '12 at 21:33

Just a hint, too long for a comment:

$$\log(\varphi(t))=\int_{0}^1 \frac{e^{i t x}-1}{x} = it + \frac{(it)^2}{2 (2!)} + \frac{(it)^3}{3(3!)} + \cdots$$

Hence, the cumulants are given by

$\kappa_1 = 1$, $\kappa_2 = \frac{1}{2}$, $\cdots$ $\kappa_n = \frac{1}{n}$

In particular: $\mu_x=1$, $\sigma^2_x=1/2$.

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If you have a compound Poisson random variable of rate $\lambda$ and jump distribution $P$, then the the characteristic function is

$$\varphi (t) = \exp \left( \lambda \int_\mathbb{R} (e^{itx}-1) \, dP(x) \right).$$

So your distribution is a weak limit of compound Poisson random variables $X_n$ where $X_n$ has rate $\lambda_n = \int_{[1/n,1]} 1/x \, dx$ and jump distribution having density $\frac{1}{x} \mathbf{1}_{[1/n,1]}$. It's a bit like the distribution of $X_1$ where $X$ is a gamma process with jumps of size greater than 1 removed. Hopefully this should give you give you some intuition.

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