Sum $\sum _{k=0}^{\infty } \frac{(-1)^k \psi (k+1)}{\left(k-\frac{3}{2}\right)^2}$

Question

Is it possible to get a closed form for: $$\sum _{k=0}^{\infty } \frac{(-1)^k \psi (k+1)}{\left(k-\frac{3}{2}\right)^2}$$ where $\psi$ is the polygamma function?

Answer 1

Since $\psi(k+1)=H_k-\gamma$ our series is given by:

$$ -4+\frac{32}{9}\gamma+4\sum_{k\geq 2}\frac{(-1)^k\left(H_k-\gamma\right)}{(2k-3)^2}=-4+\frac{32}{9}\gamma-4\gamma K+4\sum_{n\geq 0}\frac{(-1)^n H_{n+2}}{(2n+1)^2}$$ where $K$ is the Catalan constant. The last series can be computed by evaluating:

$$S_1 = \sum_{n\geq 0}\frac{(-1)^n}{(n+1)(2n+1)^2},\qquad S_2 = \sum_{n\geq 0}\frac{(-1)^n}{(n+2)(2n+1)^2},\qquad S_3 = \sum_{n\geq 0}\frac{(-1)^n\, H_n}{(2n+1)^2}$$ through the integral identities: $$ S_1 = -\int_{0}^{1}\frac{\log(1+x^2)\log x}{x^2}\,dx, \qquad S_2 = -\int_{0}^{1}\frac{\left(x^2-\log(1+x^2)\right)\log x}{x^4}\,dx,$$ $$ S_3 = \int_{0}^{1}\frac{\log(1-x^2)\log x}{1-x^2}\,dx.$$ In particular, we have: $$ S_1 = 2K-\frac{\pi}{2}+\log 2, $$ $$ S_2 = \frac{2}{3}K-\frac{\pi}{18}+\frac{1-\log 2}{9}, $$ $$ S_3 = \frac{7}{4}\zeta(3)-\frac{\pi^2}{4}\log 2.$$