"I am Back" says Integration Part 8

Substitute \begin{array} {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: $\dfrac{1}{4}\int_0^\infty \ln(t)t^{n/2-1} e^{-t} \text{dt}$ 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=n/2 and divide by four $\dfrac{1}{4}\Gamma'(n/2)=\dfrac{1}{4} \int_0^\infty \ln(t)t^{n/2-1} e^{-t} \text{dt}$ We have $\dfrac{1}{4}\Gamma'(n/2)=\dfrac{1}{4}\dfrac{\Gamma'(n/2)}{\Gamma(n/2)}\Gamma(n/2)=\dfrac{1}{4}\psi(n/2)\Gamma(n/2)$