Laplace Transform of $f(x)=\psi x^{m-1}e^{-\beta \ln^2(x)}$. I was working on some problem for my research and I came to need the Laplace transform of the following expression (which is a probability density function):
$$
f(x)=\psi x^{m-1}e^{-\beta \ln^2(x)}
$$
I thought of using taylor series for the exponential function as a whole or the $ln^2(x)$ but both gave me dumb expressions. I also tried to make a change of variable $y=\beta \ln^2(x)$ in the laplace transform but this resulted in a term that looks like $e^{-e^y}$, which I think made things even worse.
Any idea how to proceed?.
Thank you.
 A: It is possible to express the Laplace transform
$$I(s)=\int_0^\infty x^{m-1}e^{-sx-\beta \ln^2(x)}\,dx$$
as a series, by changing, as you tried, $x=e^t$:
$$I(s)=\int_{-\infty}^\infty \exp\left( -se^t-\beta t^2+mt \right)\,dt$$
Next, completing the square and shifting the variable $u=t-m/2\beta$, one obtains
\begin{align}
I(s)&=e^{\frac{m^2}{4\beta}}\int_{-\infty}^\infty\exp\left( -se^t-\beta (t-\frac{m}{2\beta})^2 \right)\,dt\\
&=e^{\frac{m^2}{4\beta}}\int_{-\infty}^\infty\exp\left( -\sigma e^u-\beta u^2 \right)\,du
\end{align}
where $\sigma=se^{m/2\beta}$. With the Taylor series of $\exp\left( -\sigma e^u \right)$ this expression becomes
$$I(s)=e^{\frac{m^2}{4\beta}}\sum_{n=0}^\infty\frac{\left( -1 \right)^n}{n!}\sigma^n
\int_{-\infty}^\infty\exp\left( nu-\beta u^2 \right)\,du$$
Completing once again the square in the exponent and shifting the variable $x=u-n/2\beta$, one obtains
\begin{align}
I(s)&=e^{\frac{m^2}{4\beta}}\sum_{n=0}^\infty\frac{\left( -1 \right)^n}{n!}\sigma^n
e^{\frac{n^2}{4\beta}}\int_{-\infty}^\infty e^{-\beta x^2}\,dx\\
&=\sqrt{\frac{\pi}{\beta}}e^{\frac{m^2}{4\beta}}\sum_{n=0}^\infty\frac{\left( -1 \right)^n}{n!}\sigma^n e^{\frac{n^2}{4\beta}}
\end{align}
Finally, the result can be written as
\begin{equation}
I(s)=\sqrt{\frac{\pi}{\beta}}\sum_{n=0}^\infty\frac{\left( -1 \right)^n}{n!}e^{\frac{(m+n)^2}{4\beta}}s^n
\end{equation} 
