Question

$\color{red} \displaystyle \rm\sum_{n = 0}^{ \infty } \frac{ \Gamma(n + 1) \Gamma(m + 1)}{(n + 1) \Gamma(n + m + 2)} = {\Psi }^{ (1) } (m + 1)$​

Answer 1

$\color{red} \displaystyle \rm\sum_{n = 0}^{ \infty } \frac{ \Gamma(n + 1) \Gamma(m + 1)}{(n + 1) \Gamma(n + m + 2)}$

$\color{red} \displaystyle \rm\sum_{n = 0}^{ \infty } \frac{ \Beta (n + 1,m + 1)}{n + 1}$

$\color{red} \displaystyle \rm\sum_{n = 0}^{ \infty } \frac{1}{n + 1} \int_{0}^{1} {t}^{n} ( 1 - {t})^{m} \: dt$

$\color{red} \displaystyle \rm \int_{0}^{1} ( 1 - {t})^{m} \sum_{n = 0}^{ \infty} \frac{ {t}^{n} }{n + 1} \: dt$

$\color{red} \displaystyle \rm \int_{0}^{1} ( 1 - {t})^{m} \sum_{n = 1}^{ \infty} \frac{ {t}^{n- 1} }{n } \: dt$

$\color{red} \displaystyle \rm - \int_{0}^{1} \frac{ ( 1 - {t})^{m} \ln(1 - t) }{t} \: dt$

$\color{red} \displaystyle \rm - \int_{0}^{1} \frac{ {t}^{m} \ln(t) }{1 - t} \: dt$

$\rm \red{ {\Psi }^{ (1) } (m + 1)}$