"I am Back" says Integration Part 7

Substitute $\begin{array}{c} {10}e^{2x}=t\to x=\dfrac{\ln(t)}{2}\\ dt= 2e^{2x} dt\\ \mid_{-\infty}^\infty \to \mid_0^\infty\end{array}$ The integrand becomes: $\dfrac{1}{4}\int_0^\infty \ln(t)t^{1/2} e^{-t} \text{dt}$ Consider the gamma functions: $\Gamma(a)=\int_0^\infty t^{a-1}e^{-t}\text{dt}$ d.w.r.a $\Gamma'(a)=\int_0^\infty \ln(t) t^{a-1} e^{-t}\text{dt}$ put a=3/2 to get that $\Gamma'(3/2)=\psi(3/2)\Gamma(3/2)=\left(\dfrac{1}{1/2}+\psi(1/2)\right)\dfrac{1}{2}\Gamma(1/2)=\dfrac{\pi^{1/2}}{2} (2-\gamma-2\ln(2))$ we know $\psi(1/2)=-\gamma+\int_0^1 \dfrac{1-x^{-1/2}}{1-x}\text{dx}=-\gamma-2\ln(2)$ final answer is $\int_{-\infty}^{\infty} {xe^{3x}e^{-{e}^{2x}} \, dx}=\dfrac{\pi^{1/2}}{8} (2-\gamma-2\ln(2))$