Given the distribution $$n(r) = \frac{1}{\sqrt{2\pi}rln\sigma_g}exp^{\frac{-ln^2(r/r_g)}{2ln^2(\sigma_g)}}$$ Where $r$&$r_g$ have the same physical units and $\sigma_g$ is a dimensionless parameter. My task is to find the $v^{th}$ moment by definition.
So far, I think my integration limits are from $0$ to $\infty$. Furthermore, I think a change of variable could be used such that $x = ln(\frac{r}{r_g})$. However, even with the change in variable, it seems to leave two variables in the integration - thus I'm not sure whether to treat r like a constant once we replace with $dx$. Here's my work so far: $$E(r^v) = \int_{0}^{\infty}(r^v*\frac{1}{\sqrt{2\pi}rln\sigma_g}e^{\frac{-ln^2(r/r_g)}{2ln^2(\sigma_g)}}) dr$$ $$x = ln(r/r_g)$$ $$rdx = dr$$ $$E(r^v) = \int_{0}^{\infty}(r^v*\frac{1}{\sqrt{2\pi}rln\sigma_g}e^{\frac{-x^2}{2ln^2(\sigma_g)}}) dx * (r)$$ $$E(r^v) = \int_{0}^{\infty}(r^v*\frac{1}{\sqrt{2\pi}ln\sigma_g}e^{\frac{-x^2}{2ln^2(\sigma_g)}})dx$$ $$E(r^v) =r^v*\frac{1}{\sqrt{2\pi}rln\sigma_g} \int_{0}^{\infty}(e^{\frac{-x^2}{2ln^2(\sigma_g)}}) dx$$
I'm stumped and need help solving this integral. Much thanks!
The last integral can be solved by taking a change of variable $t=\frac{x^2}{2\ln^2\sigma_g}$ and then using the gamma function to solve it, being: $dx=x^{-1}\ln^2{\sigma_g}dt$ and $x=\sqrt{2}\ln{\sigma_g}\sqrt t$, the integral takes the next form:
$\int_0^{\infty}e^{\frac{-x^2}{2\ln^2\sigma_g}}dx=\int_0^{\infty}\frac{\ln^2\sigma_g}{\sqrt 2\ln\sigma_g}t^{-\frac{1}{2}}e^{-t}dt=\frac{\ln\sigma_g}{\sqrt 2}\Gamma(\frac{1}{2})=\frac{\sqrt \pi\ln\sigma_g}{\sqrt 2}$
So you just substitute this for your integral in the last step and get the final result.
However, I am not sure if you are doing the integration correctly, as you are taking the $r$ variable outside the integral after the change of variable, but those $r$ are variables that depend on $x$, and so you cannot take them out of the integral. You should write all in terms of $x$ so that you can solve for x.