1. ## Computing an Integral

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

2. Originally Posted by Arka

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

That's easy:

$\displaystyle \int_0^z\ln\Gamma(x+1)dx=\frac{z}{2}\ln(2\pi)-\frac{z(z+1)}{2}+z\ln\Gamma(z+1)$ $\displaystyle -\ln\left\{(2\pi)^{\frac{z}{2}}\exp\left(\frac{-z(z+1)}{2}-\frac{\gamma z^2}{2}\right)\prod_{k=1}^{\infty}\left\{\left(1+\ frac{z}{k}\right)^k\exp\left(-z+\frac{z^2}{2k}\right)\right\}\right\}$

3. Originally Posted by Arka

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

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...

$\displaystyle \ln \Gamma (1+x)= - \gamma\ x + \sum_{n=2}^{\infty} (-1)^{n} \zeta(n)\ \frac{x^{n}}{n}$ (1)

... where $\gamma$ is the 'Euler's constant' and $\zeta (*)$ the 'Riemann zeta function'... integrating (1) 'term by term' You obtain...

$\displaystyle \int_{0}^{z} \ln \Gamma (1+x)\ dx= - \gamma\ \frac{z^{2}}{2} + \sum_{n=2}^{\infty} (-1)^{n} \zeta(n)\ \frac{z^{n+1}}{n\ (n+1)}$ (2)

Kind regards

$\chi$ $\sigma$

4. Originally Posted by Drexel28
That's easy:

$\displaystyle \int_0^z\ln\Gamma(x+1)dx=\frac{z}{2}\ln(2\pi)-\frac{z(z+1)}{2}+z\ln\Gamma(z+1)$ $\displaystyle -\ln\left\{(2\pi)^{\frac{z}{2}}\exp\left(\frac{-z(z+1)}{2}-\frac{\gamma z^2}{2}\right)\prod_{k=1}^{\infty}\left\{\left(1+\ frac{z}{k}\right)^k\exp\left(-z+\frac{z^2}{2k}\right)\right\}\right\}$
Would you please show me the steps?........because I couldn't really get you.