"I am Back" says Integration Part 6

Substitute $\begin{array}{c}10e^{2x}=t\to x=\dfrac{\ln(t)}{2}\\ dt= 2e^{2x} dt\\ \mid_{-\infty}^\infty \to \mid_0^\infty\end{array}$ The integrand becomes: $\dfrac{1}{4}\int_0^\infty \ln(t) e^{-t} \text{dt}$ 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=1 to get our original integrand. Now $\Gamma'(1)=\psi(1)\Gamma(1)=-\gamma$ so the answer is $\dfrac{-\gamma}{\boxed{4}}$