Infinite Series :$ \sum_{n=0}^\infty \frac{\Gamma \left(n+\frac{1}{2} \right)\psi \left(n+\frac{1}{2} \right)}{n! \left(n+\frac{3}{2}\right)^2}$ Prove that:
$$\sum_{n=0}^\infty \frac{\Gamma \left(n+\frac{1}{2} \right)\psi \left(n+\frac{1}{2} \right)}{n! \left(n+\frac{3}{2}\right)^2} = \frac{-\pi^{\frac{3}{2}}}{12}\left( \pi^2+6\gamma(1-2\log 2)-12\log 2\right)$$
where $\gamma$ is Euler-Mascheroni Constant and $\psi(z)$ is the Digamma Function. 
 A: Consider the series
\begin{align}
S(x,y) = \sum_{n=0}^{\infty} \frac{\Gamma(n+x)}{n! \ (n+y+1)^{2}}.
\end{align}
This series can be evaluated as follows
\begin{align}
S(x,y) &= \sum_{n=0}^{\infty} \frac{\Gamma(x) \ (x)_{n}}{\Gamma(2) \ n!} \ \int_{0}^{\infty} e^{-(n+y+1)t} \ t \ dt \\
&= \Gamma(x) \ \int_{0}^{\infty} e^{-(y+1)t} \left( \sum_{n=0}^{\infty} \frac{\Gamma(x) \ (x)_{n} \ e^{-nt}}{n!} \right) \ t \ dt \\
&= \Gamma(x) \ \int_{0}^{\infty} e^{-(y+1)t} \ t \ (1-e^{-t})^{-x} \ dt \\
&= - \Gamma(x) \ \int_{0}^{1} u^{y} (1-u)^{-x} \ \ln(u) \ du \\
&= - \Gamma(x) \partial_{y} B(y+1, 1-x) \\
S(x,y) &= - \Gamma(x) \ B(y+1, 1-x) \ \left[ \psi(y+1) - \psi(2+y-x) \right].
\end{align}
which yields
\begin{align}
\sum_{n=0}^{\infty} \frac{\Gamma(n+x)}{n! \ (n+y+1)^{2}} = - \Gamma(x) \ B(y+1, 1-x) \ \left[ \psi(y+1) - \psi(2+y-x) \right].
\end{align}
Taking the derivative with respect to $x$ leads to 
\begin{align}
\partial_{x} S(x,y) &= - \Gamma(x) \ B(y+1,1-x) \ \left[ \psi_{1}(2+y-x) + (\psi(y+1) - \psi(2+y-x)) \right. \\
& \hspace{15mm} \left. \cdot(\psi(2+y-x) - \psi(1-x) + \psi(x)) \right] 
\end{align} 
which yields
\begin{align}
\sum_{n=0}^{\infty} \frac{\Gamma(n+x) \ \psi(n+x)}{n! \ (n+y+1)^{2}} &=
- \Gamma(x) \ B(y+1,1-x) \ \left[ \psi_{1}(2+y-x) \right. \\
& \hspace{15mm} \left. + (\psi(y+1) - \psi(2+y-x)) \cdot(\psi(2+y-x) - \psi(1-x) + \psi(x)) \right]. 
\end{align}
Taking $x=y=1/2$ leads to the two series
\begin{align}
\sum_{n=0}^{\infty} \frac{\Gamma\left(n+\frac{1}{2} \right)}{n! \ \left(n+\frac{3}{2}\right)^{2}} =  \frac{\pi^{3/2}}{2} ( 2 \ln 2 -1).
\end{align}
and
\begin{align}
\sum_{n=0}^{\infty} \frac{\Gamma\left(n+\frac{1}{2}\right) \ \psi\left(n+\frac{1}{2}\right)}{n! \ \left(n+ \frac{3}{2} \right)^{2}} &=
- \frac{\pi^{3/2}}{12} \ \left[ \pi^{2} + 6 \gamma (1 - 2 \ln 2 ) - 12 \ln 2 \right]. 
\end{align}
A: Begin with
$$\sum_{n=0}^\infty \frac{\Gamma(n+1-y)}{n!}\frac{1}{x+n} = \frac{\pi \Gamma(x)}{\sin(\pi y) \Gamma(x+y)}$$
Differentiating with respect to $x$ gives
$$-\sum_{n=0}^\infty \frac{\Gamma(n+1-y)}{n!}\frac{1}{(x+n)^2} = \frac{\pi \Gamma(x)}{\sin(\pi y) \Gamma(x+y)} \left\{ \psi(x)-\psi(x+y)\right\}$$
Now, differentiate with respect to $y$:
$$\sum_{n=0}^\infty \frac{\Gamma(n+1-y)\psi(n+1-y)}{n!}\frac{1}{(n+x)^2}=\frac{-\pi \Gamma(x)}{\sin(\pi y)\Gamma(x+y)}\left[ \{ \sin(\pi y)\psi(x+y)+\pi\cos(\pi y)\}\{\psi(x)-\psi(x+y)\}+\psi_1(x+y)\right]$$
Putting $x=\frac{3}{2}$ and $y=\frac{1}{2}$, gives the desired result.
$$\sum_{n=0}^\infty \frac{\Gamma \left(n+\frac{1}{2} \right)\psi \left(n+\frac{1}{2} \right)}{n! \left(n+\frac{3}{2}\right)^2} = \frac{-\pi^{\frac{3}{2}}}{12}\left( \pi^2+6\gamma(1-2\log 2)-12\log 2\right)$$
If you have any other method, please enlighten me.
