What is $\int_0^1{\ln\Gamma(x)\sin(\pi x)}\mathop{\mathrm{d}x}$?

Question

Hello I am stuck with this integral:

$$\int_0^1{\ln\Gamma(x)\sin(\pi x)}\mathop{\mathrm{d}x}$$ My questions are:

What is the integrand, $\ln(\Gamma(x)\sin(x\pi))$ or $\ln(\Gamma(x))\:\sin(x\pi)$?
Are there any other suggestions?

Thank you for your help.

Probably the former. If it were the latter you would just have been given ln$\Gamma(x)+$ ln$($sin$(x))$ to integrate. — Arkady, Jun 16 '16 at 02:44
Usually the notation for log gamma is just written as $\ln\Gamma$, so your second one is prolly the thing you should be concerned about. (@Ravi's comment momentarily confused me.) — J. M. ain't a mathematician, Jun 16 '16 at 02:57
It is very likely the second one. Using $\Gamma(x)\Gamma(1-x) = \frac{\pi}{\sin(\pi x)}$, one can obtain a closed form of the integral $\frac{1+\log(\pi/2)}{\pi}$. I'm not sure whether the first version has a closed form expression or not. — achille hui, Jun 16 '16 at 03:19
The first one for sure has a closed form $\int_{0}^{1} \ln (\Gamma (x)) dx=\frac{1}{2} \ln(2\pi)$ is in Wikipedia. @achillehui — Ahmed S. Attaalla, Jun 16 '16 at 03:29
Related: http://math.stackexchange.com/questions/784529/integral-int-01-log-left-gamma-leftx-alpha-right-right-rm-dx-frac and https://en.wikipedia.org/wiki/Joseph_Ludwig_Raabe — Jack D'Aurizio, Jun 16 '16 at 04:19
@J.M. Yes, I really did mean latter. Sorry for the confusion. — Arkady, Jun 16 '16 at 04:38
Always use $\left(\right)$'s. For example: $\color{#f00}{\ln\left(2\right)}$ instead of $\ln 2$. — Felix Marin, Jun 16 '16 at 19:55

score 4 · Accepted Answer · answered Jun 16 '16 at 04:37

Let $I_1$ and $I_2$ be the integrals

$$ I_1 \stackrel{def}{=} \int_0^1 \log(\Gamma(x)\sin(\pi x)) dx \quad\text{ and }\quad I_2 \stackrel{def}{=} \int_0^1 \log(\Gamma(x))\sin(\pi x) dx $$ I believe $I_2$ is the integral intended. Notice $$\sin\pi x = \sin \pi(1-x)\quad\text{ and }\quad\Gamma(x)\Gamma(1-x) = \frac{\pi}{\sin \pi x}$$ We have $$ I_2 = \frac12\int_0^1 \left(\log\Gamma(x) + \log\Gamma(1-x)\right)\sin\pi x dx = \frac12\int_0^1 \log\left(\frac{\pi}{\sin\pi x}\right)\sin \pi x dx $$ Change variable first to $y = \pi x$ and then to $z = \cos y$, we have

$$\begin{align} I_2 &= \frac{1}{2\pi}\int_0^{\pi}\left(\log\pi - \log\sin y\right)\sin y dy = \frac{1}{\pi}\int_0^{\pi/2}\left(\log\pi - \log\sin y\right)\sin y dy\\ &= \frac{1}{\pi}\left(\log \pi - \int_0^1 \log \sqrt{1-z^2} dz\right)\\ &= \frac{1}{\pi}\left(\log\pi - \frac12\bigg[(1+z)\log(1+z)-(1-z)\log(1-z)-2z\bigg]_0^1\right)\\ &= \frac{1+\log(\pi/2)}{\pi} \end{align} $$ For completeness, let us evaluate $I_1$ too. By a similar argument like above, we have $$I_1 = \frac{1}{\pi}\int_0^{\pi/2}(\log \pi + \log\sin y)dy$$ Notice $$\int_0^{\pi/2}\log\sin(2y) dy = \frac12 \int_0^{\pi}\log\sin y dy = \int_0^{\pi/2}\log\sin y dy$$ We have $$\int_0^{\pi/2}\log\sin y dy = \int_0^{\pi/2}\log\cos y dy = \int_0^{\pi/2}(\log\sin(2y) - \log\sin y - \log 2) dy = -\log 2\frac{\pi}{2}$$ This implies $$I_1 = \frac{1}{\pi}\left(\log\pi\cdot\frac{\pi}{2} - \log 2\cdot\frac{\pi}{2}\right) = \frac12\log\frac{\pi}{2}$$ the result first pointed by @J.M.

Jack D'Aurizio · Answer 2 · 2016-06-16T15:14:58.143

$$\begin{eqnarray*} \int_{0}^{1}\log\Gamma(x)\,\sin(\pi x)\,dx &=& \log(\pi)\int_{0}^{1}(1-x)\sin(\pi x)\,dx-\frac{1}{2}\int_{0}^{1}\sin(\pi x)\log\sin(\pi x)\,dx\\&=&\frac{\log \pi}{\pi}-\frac{\log(2)-1}{\pi}=\color{red}{\frac{1}{\pi}\left(1+\log\frac{\pi}{2}\right)}\end{eqnarray*}$$

can be directly computed from the Kummer's Fourier series expansion for $\log\Gamma$.

A simple alternative is given by integration by parts and the reflection formula for the $\psi$ function:

$$\begin{eqnarray*} \pi I = -\int_{0}^{1}\psi(x)(1-\cos(\pi x))\,dx.\end{eqnarray*}$$

What is $\int_0^1{\ln\Gamma(x)\sin(\pi x)}\mathop{\mathrm{d}x}$?

2 Answers2