Integration – Evaluating the Log Gamma Integral in Terms of the Hurwitz Zeta Function

gamma functionintegrationriemann-zetaspecial functionszeta-functions

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?

Best Answer

To evaluate that limit, we can expand each function in a Laurent series at $s=0$ and use the following 3 facts about the Hurwitz zeta function:

$$ \zeta(-s,a) = \zeta(-s,a+1) + a^{s} \tag{1}$$

$$ \zeta'(-s,a) = \zeta'(-s,a+1) -a^{s} \log(a) $$

$$\zeta(-n, a) = -\frac{B_{n+1}(a)}{n+1} \ , \ n \in\mathbb{N} \tag{2}$$

Doing so, we get

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

$$+ \Gamma(s-1) \zeta(s-1) \Big]$$

$$ = z - z \log z - \frac{\gamma z^{2}}{2}$$

$$ + \lim_{s \to 0^{+}} \Bigg[-\Big(-\frac{1}{s} + \gamma -1 + \mathcal{O}(s) \Big) \Big( -\frac{z^{2}}{2}+\frac{z}{2}-\frac{1}{12}-z + \zeta'(-1,z)s + z \log z \ s + \mathcal{O}(s^{2}) \Big)$$

$$ - z \Big( \frac{1}{s} - \gamma + \mathcal{O}(s) \Big) \Big( - \frac{1}{2} - \frac{\log (2 \pi)}{2} s + \mathcal{O}(s^{2}) \Big) + \frac{z^{2}}{2} \Big(1- \gamma s + \mathcal{O}(s^{2}) \Big) \Big( \frac{1}{s} + \gamma + \mathcal{O} (s) \Big) $$

$$ + \Big(- \frac{1}{s} + \gamma -1 + \mathcal{O} (s) \Big) \Big( - \frac{1}{12} + \zeta'(-1) s + \mathcal{O}(s^{2}) \Big) \Bigg] $$

$$ = z - z \log z - \frac{\gamma z^{2}}{2}$$

$$ + \lim_{s \to 0^{+}} \Big[\zeta'(-1,z) + z \log z + \frac{\gamma z^{2}}{2} - \frac{z^{2}}{2} + \frac{z}{2} - z + \frac{z \log(2 \pi)}{2} - \zeta(-1)+ \mathcal{O}(s) \Big] $$

$$ = \frac{z}{2} \log(2 \pi) + \frac{z(1-z)}{2} -\zeta'(-1)+ \zeta'(-1,z) $$

$ $

$(1)$ http://dlmf.nist.gov/25.11 (25.11.3)

$(2)$ http://mathworld.wolfram.com/HurwitzZetaFunction.html (9)

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