13
$\begingroup$

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.

$\endgroup$
11
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.