One way to evaluate ∫z0logΓ(x)dx is in terms of the Barnes G-function.
∫z0logΓ(x)dx=z2log(2π)+z(1−z)2+zlogΓ(z)−logG(z+1)
Another way is in terms of the Hurwitz zeta function.
∫z0logΓ(x)dx=z2log(2π)+z(1−z)2−ζ′(−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)xk−1=ψ(a)−ψ(a−x).
Integrating both sides, I get ∞∑k=2ζ(k,a)kxk=ψ(a)x+logΓ(a−x)−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)∫∞0ts−1e−at1−e−tdt, I get
∫z0logΓ(x)dx=z−zlogz−γz22+∞∑k=2(−1)kzk+1k(k+1)1Γ(k)∫∞0tk−1e−t1−e−tdt
=z−logz−γz22+z∫∞0e−t1−e−t1t∞∑k=2(−1)kk+1(zt)kk!dt
=z−zlogz−γz22+z∫∞0e−t1−e−t1t(−e−ztzt−1+zt2+1zt)dt
=z−zlogz−γz22+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)
Attribution
Source : Link , Question Author : Random Variable , Answer Author : Random Variable