# Digamma function in expectation

I have the moment generating function $$M\left(t\right)=\frac{\Gamma\left(\alpha+1\right)\Gamma\left(1-t\right)}{\Gamma\left(\alpha-t+1\right)},\ t<-1.$$ We know that expectation and variance can be found by $E\left(X\right)=\left.\frac{d\ln M\left(t\right)}{dt}\right|_{t=0}$ and $Var\left(X\right)=\left.\frac{d^{2}\ln M\left(t\right)}{dt^{2}}\right|_{t=0}$. How to show that $$E\left(X\right)=\psi\left(\alpha+1\right)-\psi\left(1\right)\ \text{and}\ Var\left(X\right)=\psi'\left(1\right)-\psi'\left(\alpha+1\right)$$ where $\psi\left(x\right)=\frac{d}{dx}\ln\Gamma\left(x\right)$ is digamma function.

at first, it seems obvious but; i couldn't get $\psi\left(\alpha+1\right)$ term in expectation.

$$\ln M\left(t\right)=\ln\Gamma\left(\alpha+1\right)+\ln\Gamma\left(1-t\right)-\ln\Gamma\left(\alpha-t+1\right).$$ Hence $$\frac{d}{dt}\ln M(t)=\frac{d}{dt}\ln\Gamma\left(1-t\right)-\frac{d}{dt}\ln\Gamma\left(\alpha-t+1\right)=-\psi(1-t)+\psi(\alpha-t+1),$$ and at $t=0$ you get $\psi(\alpha+1)-\psi(1)$.