1. Computing an Integral

$\displaystyle \int_0^z\ln\Gamma(x+1)dx$

$\displaystyle \int_0^z\ln\Gamma(x+1)dx$
$\displaystyle \displaystyle \int_0^z\ln\Gamma(x+1)dx=\frac{z}{2}\ln(2\pi)-\frac{z(z+1)}{2}+z\ln\Gamma(z+1)$$\displaystyle \displaystyle -\ln\left\{(2\pi)^{\frac{z}{2}}\exp\left(\frac{-z(z+1)}{2}-\frac{\gamma z^2}{2}\right)\prod_{k=1}^{\infty}\left\{\left(1+\ frac{z}{k}\right)^k\exp\left(-z+\frac{z^2}{2k}\right)\right\}\right\} 3. Originally Posted by Arka I cant figure out this integral............ please help \displaystyle \int_0^z\ln\Gamma(x+1)dx Thanks in advance.. In a previous post [I don't remember exactly ... but I think is in the 'math challenge forum'...] there is a [non elementar..] expansions I found some years ago... \displaystyle \displaystyle \ln \Gamma (1+x)= - \gamma\ x + \sum_{n=2}^{\infty} (-1)^{n} \zeta(n)\ \frac{x^{n}}{n} (1) ... where \displaystyle \gamma is the 'Euler's constant' and \displaystyle \zeta (*) the 'Riemann zeta function'... integrating (1) 'term by term' You obtain... \displaystyle \displaystyle \int_{0}^{z} \ln \Gamma (1+x)\ dx= - \gamma\ \frac{z^{2}}{2} + \sum_{n=2}^{\infty} (-1)^{n} \zeta(n)\ \frac{z^{n+1}}{n\ (n+1)} (2) Kind regards \displaystyle \chi \displaystyle \sigma 4. Originally Posted by Drexel28 That's easy: \displaystyle \displaystyle \int_0^z\ln\Gamma(x+1)dx=\frac{z}{2}\ln(2\pi)-\frac{z(z+1)}{2}+z\ln\Gamma(z+1)$$\displaystyle \displaystyle -\ln\left\{(2\pi)^{\frac{z}{2}}\exp\left(\frac{-z(z+1)}{2}-\frac{\gamma z^2}{2}\right)\prod_{k=1}^{\infty}\left\{\left(1+\ frac{z}{k}\right)^k\exp\left(-z+\frac{z^2}{2k}\right)\right\}\right\}$