Evaluating the log gamma integral ∫z0logΓ(x)dx\int_{0}^{z} \log \Gamma (x) \, \mathrm dx in terms of the Hurwitz zeta function

One way to evaluate $\displaystyle\int_{0}^{z} \log \Gamma(x) \, \mathrm dx$ is in terms of the Barnes G-function.

$$\int_{0}^{z} \log \Gamma(x) \, \mathrm dx = \frac{z}{2} \log (2 \pi) + \frac{z(1-z)}{2} + z \log \Gamma(z) - \log G(z+1)$$

Another way is in terms of the Hurwitz zeta function.

$$\int_{0}^{z} \log \Gamma(x) \, \mathrm dx = \frac{z}{2} \log(2 \pi) + \frac{z(1-z)}{2} - \zeta^{'}(-1) + \zeta^{'}(-1,z)$$

I’ve been trying to prove the latter so that I can prove $$\log G(z+1) - z \log \Gamma(z) = \zeta'(-1) - \zeta'(-1,z) .$$

My starting point is the generating function $$\sum_{k=2}^{\infty} \zeta(k,a) x^{k-1} = \psi(a) - \psi(a-x) .$$

Integrating both sides, I get $$\sum_{k=2}^{\infty} \frac{\zeta(k,a)}{k} x^{k} = \psi(a) x + \log \Gamma(a-x) - \log \Gamma(a),$$

which implies

$$\sum_{k=2}^{\infty} (-1)^{k} \frac{\zeta(k,1)}{k} x^{k} = \gamma x + \log \Gamma(x+1) .$$

Then rearranging and integrating both sides from $0$ to $z$, I get

$$\int_{0}^{z} \log \Gamma(x+1) \, \mathrm dx = \int_{0}^{z} \log x \ \mathrm dx + \int_{0}^{z} \log \Gamma(x) \, \mathrm dx = - \frac{\gamma z^{2}}{2} + \sum_{k=2}^{\infty} (-1)^{k} \frac{\zeta(k,1)}{k(k+1)} z^{k+1} .$$

And then using the integral representation $$\zeta(s,a) = \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{t^{s-1} e^{-at}}{1-e^{-t}} \, \mathrm dt,$$ I get

$$\int_{0}^{z} \log \Gamma(x) \, \mathrm dx = z- z \log z - \frac{\gamma z^{2}}{2} + \sum_{k=2}^{\infty} (-1)^{k} \frac{z^{k+1}}{k(k+1)} \frac{1}{\Gamma(k)} \int_{0}^{\infty} \frac{t^{k-1} e^{-t}}{1-e^{-t}} \, \mathrm dt$$

$$= z - \log z - \frac{\gamma z^{2}}{2} + z \int_{0}^{\infty} \frac{e^{-t}}{1-e^{-t}} \frac{1}{t} \sum_{k=2}^{\infty} \frac{(-1)^{k}}{k+1} \frac{(zt)^{k}}{k!} \, \mathrm dt$$

$$= z - z \log z - \frac{\gamma z^{2}}{2} + z \int_{0}^{\infty} \frac{e^{-t}}{1-e^{-t}}\frac{1}{t} \left( -\frac{e^{-zt}}{zt} - 1 + \frac{zt}{2} + \frac{1}{zt} \right) \, \mathrm dt$$

$$= z - z \log z - \frac{\gamma z^{2}}{2} + \lim_{s \to 0^{+}} \Big[ - \int_{0}^{\infty} \frac{t^{s -2} e^{-(z+1)t}}{1-e^{-t}} \, \mathrm dt - z \int_{0}^{\infty} \frac{t^{s-1} e^{-t}}{1-e^{-t}} \, \mathrm dt$$ $$+ \frac{z^{2}}{2} \int_{0}^{\infty} \frac{t^{s} e^{-t}}{1-e^{-t}} \, \mathrm dt + \int_{0}^{\infty} \frac{t^{s -2} e^{-t}}{1-e^{-t}} \, \mathrm dt \Big]$$

$$= z - z \log z - \frac{\gamma z^{2}}{2} + \lim_{s \to 0^{+}} \Big[ - \Gamma(s-1) \zeta(s-1,z+1) -z \Gamma(s) \zeta(s) + \frac{z^{2}}{2} \Gamma(s+1) \zeta(s+1)$$
$$+ \Gamma(s-1) \zeta(s-1) \Big] .$$

Assuming I haven’t made any mistakes up to this point, how do I evaluate that limit?

Answer

Attribution
Source : Link , Question Author : Random Variable , Answer Author : Random Variable