From the Taylor series expansion of the Gamma function, I know that:
$$ \Gamma\!\left(z + \frac12\right) = \Gamma(z) + \sum_{k=0}^\infty \frac{\Gamma^{(k)}(z)}{k!}\left(\frac12\right)^k, $$ where $\Gamma^{(k)}(z) = \frac{d^k}{dz^k}\Gamma(z)$, and for any $z$. My question is: is there an (perhaps, analytical) expression for these derivatives at $z\in\mathbb N$, a positive integer.
Thanks for any suggestions!
You can obtain a recurrence relation for those values. First note that, for $k\ge 1$, $$ \frac{{{\rm d}^k \Gamma (z)}}{{{\rm d}z^k }} = \frac{{{\rm d}^{k - 1} }}{{{\rm d}z^{k - 1} }}(\psi^{(0)} (z)\Gamma (z)) = \sum\limits_{j = 0}^{k - 1} {\binom{k-1}{j}\psi ^{(j)} (z)\frac{{{\rm d}^{k - j - 1} \Gamma (z)}}{{{\rm d}z^{k - j - 1} }}} , $$ where $\psi^{(j)}$ is a polygamma function. Assume that $z = n \ge 1$ is an integer. Then $$ \psi ^{(0)} (n) = - \gamma + \sum\limits_{r = 1}^{n - 1} {\frac{1}{r}}, $$ with $\gamma$ being the Euler–Mascheroni constant, and $$ \psi ^{(j)} (n) = ( - 1)^{j + 1} j!\sum\limits_{r = 0}^\infty {\frac{1}{{(r + n)^{j + 1} }}} = ( - 1)^{j + 1} j!\bigg( {\zeta (j + 1) - \sum\limits_{r = 1}^{n - 1} {\frac{1}{{r^{j + 1} }}} } \bigg) $$ for $j\ge 1$, with $\zeta$ being the Riemann zeta function.
Of course, the formula $$ \Gamma\! \left( {n + \frac{1}{2}} \right) = \frac{{(2n)!}}{{4^n n!}}\sqrt \pi $$ provides a much more efficient way to compute the gamma function for half-integer values.