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 z0logΓ(x)dx is in terms of the Barnes G-function.

z0logΓ(x)dx=z2log(2π)+z(1z)2+zlogΓ(z)logG(z+1)

Another way is in terms of the Hurwitz zeta function.

z0logΓ(x)dx=z2log(2π)+z(1z)2ζ(1)+ζ(1,z)

I’ve been trying to prove the latter so that I can prove logG(z+1)zlogΓ(z)=ζ(1)ζ(1,z).

My starting point is the generating function k=2ζ(k,a)xk1=ψ(a)ψ(ax).

Integrating both sides, I get k=2ζ(k,a)kxk=ψ(a)x+logΓ(ax)logΓ(a),

which implies

k=2(1)kζ(k,1)kxk=γx+logΓ(x+1).

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

z0logΓ(x+1)dx=z0logx dx+z0logΓ(x)dx=γz22+k=2(1)kζ(k,1)k(k+1)zk+1.

And then using the integral representation ζ(s,a)=1Γ(s)0ts1eat1etdt, I get

z0logΓ(x)dx=zzlogzγz22+k=2(1)kzk+1k(k+1)1Γ(k)0tk1et1etdt

=zlogzγz22+z0et1et1tk=2(1)kk+1(zt)kk!dt

=zzlogzγz22+z0et1et1t(eztzt1+zt2+1zt)dt

=zzlogzγz22+lim + \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

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

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