The question is to compute the entropy of density function $\frac{{x{e^{ - \frac{{{x^2}}}{{2t}}}}}}{{\sqrt {2\pi {t^3}} }}$.
First anyone knows what this distribution is? $x$ can only take non-negative values in this density. If $x$ ranges in $[0,\infty)$, then $t$ is solvable to be $t=\frac{1}{2\pi}$, and the density becomes $2\pi xe^{-\pi x^2}$.
Suppose $x\in[0,+\infty)$, then $ - \int_0^{ + \infty } {\frac{{x{e^{ - \frac{{{x^2}}}{{2t}}}}}}{{\sqrt {2\pi {t^3}} }}[\ln x - \frac{{{x^2}}}{{2t}} - \frac{1}{2}\ln 2\pi {t^3}]dx} = \int_0^{ + \infty } {\frac{{{x^3}{e^{ - \frac{{{x^2}}}{{2t}}}}}}{{2t\sqrt {2\pi {t^3}} }}} + \int_0^{ + \infty } {\frac{{\ln 2\pi {t^3}x{e^{ - \frac{{{x^2}}}{{2t}}}}}}{{4t\sqrt {2\pi {t^3}} }}} - \int_0^{ + \infty } {\frac{{x\ln x{e^{ - \frac{{{x^2}}}{{2t}}}}}}{{2t\sqrt {2\pi {t^3}} }}} = - \int_0^{ + \infty } {\frac{{x\ln x{e^{ - \frac{{{x^2}}}{{2t}}}}}}{{2t\sqrt {2\pi {t^3}} }}} $.
Then I got stuck here. Any one knows how to compute the integral $\int_0^{ + \infty } {x\ln x{e^{ - \frac{{{x^2}}}{{2t}}}}} $? Thanks!
First question: This is as Rayleigh distribution
which can be interpreted as the density $\rho(x)$ of the length $x=\sqrt{x_1^2+x_2^2}$ of a vector $x=(x_1,x_2)$ with independent, gaussian distributed components.
Second question:
After clearing up the notation and trivial rescaling $x\rightarrow x/\sqrt{2 t}$ we want to calculate (there is indeed another part without $\log(x)$ which is easy)
$$ I=\int_0^{\infty}\log(x)xe^{-x^2} $$
we observe that
$$ I=\partial_a \int_0^{\infty}x^{a+1}e^{-x^2}\big|_{a=0}=\partial_a J(a)\big|_{a=0} $$
$J(a)$ is readily expressed in terms of gamma functions: $$ J(a)=\frac{1}{2}\Gamma \left(\frac{a}{2}+1\right) $$
and therefore $$ I=\frac{1}{4}\Gamma(\frac{a}{2}+1)\psi(\frac{a}{2}+1)\big|_{a=0}=\frac{1}{4}\cdot 1\cdot(-\gamma)=\frac{-\gamma}{4} $$ where we used the special value $\psi(1)=-\gamma$ of the Digamma function
Edit: This is indeed one of the integral reps. of the Euler-Marschoni constant, given on wikipedia