# Distribution with $n$'th cumulant given by $\frac 1 n$?

Is there any information out there about the distribution whose $n$'th cumulant is given by $\frac 1 n$? The cumulant-generating function is of the form $$\kappa(t) = \int_0 ^ 1 \frac{e^{tx} - 1}{x} \ dx.$$ I've run across it as the limiting distribution of some random variables but I haven't been able to find any information on it.

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I cannot see that this function $\kappa(t)$ you have given has the claimed property! You should revise yoiur work. Approximating the exponential n the integrand close to zero with $1+tx$, the integrand close to zero becomes $t/x$, so is divergent. So that integral cannot represent a cumulant generating function. –  kjetil b halvorsen Jul 11 at 15:32
@kjetilbhalvorsen not sure I follow. Approximating $e^{tx}$ with $1 + tx$ gives $\frac{tx}{x} = t$ for the integrand. Also, according to this the function I gave has a known integral in terms of hyperbolic cosine and sine integrals. To show that $\kappa(t)$ has the claimed property just do a full Taylor series around $0$ for $e^{tx}$ and push the integral through to sum to get the Taylor series for $\kappa(t)$ around $0$. –  guy Jul 11 at 15:54
sympy says the integral is divergent (in its own eccentric way!). But sympy must be wrong, I see it now, experimented with some numerical integration, and it works just well. Will try again. –  kjetil b halvorsen Jul 11 at 16:36
Looking at Wolphram alphas result, it cannot be correct either, it haves a non-zero limit when t approaches zero, while $\kappa(0)=0$ clearly. –  kjetil b halvorsen Jul 11 at 19:10
@kjetilbhalvorsen It does have zero as a limit. I promise this is a cgf. –  guy Jul 11 at 19:19