Show that $\int_0^1(\ln\Gamma)(x)\mathrm dx=\ln(\sqrt{2\pi})$

Im totally stuck with this exercise. It is supposed that I must solve it using the reflection formula for the Gamma function. My work so far:

Using the reflection formula we have

$$\int_0^1(\ln \Gamma)(z)\mathrm dz=-\int_0^1(\ln\Gamma)(1-z)\mathrm dz-\int_0^1\ln\left(\frac{\sin(\pi z)}\pi\right)\mathrm dz=\\=-\int_0^1(\ln\Gamma)(z)\mathrm dz-\int_0^1\ln\left(\frac{\sin(\pi z)}\pi\right)\mathrm dz$$


$$\int_0^1(\ln \Gamma)(z)\mathrm dz=-\frac12\int_0^1\ln\left(\frac{\sin(\pi z)}\pi\right)\mathrm dz=\frac{\ln \pi}2-\frac12\int_0^1(\ln \sin)(\pi z)\mathrm dz$$

Then it must be the case that

$$\int_0^1(\ln\sin)(\pi z)\mathrm dz=-\ln 2$$

but I dont know how to solve this last step.

Some of the identities at my disposition are:

$$\sin(\pi z)=\pi z\prod_{k=1}^\infty\left(1-\frac{z^2}{k^2}\right),\; z\in\Bbb C\quad\quad\Gamma\left(\frac{x}2\right)\Gamma\left(\frac{x+1}2\right)=\frac{\sqrt\pi}{2^{x-1}}\Gamma(x),\; x\in(0,\infty)$$

$$(\ln \Gamma)(1+z)=-\gamma z+\sum_{k=2}^\infty(-1)^k\frac{\zeta(k)}kz^k,\, |z|<1\quad\quad\zeta(2k)=\frac{(-1)^{k+1}(2\pi)^{2k}}{2(2k)!}B_{2k},\, k\in\Bbb N_{>0}$$

$$\pi z\cot(\pi z)=1+2z^2\sum_{k=0}^\infty\frac1{z^2-n^2},\;z\in\Bbb C\setminus\Bbb Z$$

And for $z\in\Bbb C{\setminus}({-}\Bbb N)$:

$$\frac1{\Gamma(z)}=ze^{\gamma z}\prod_{k=1}^\infty\left(1+\frac{z}k\right)e^{-z/k}$$


where $\gamma$ is the Euler-Mascheroni constant and $B_{2k}$ are the Bernoulli numbers.

From above, if there is no weird mistake somewhere, I get the identities

$$\int_0^1(\ln\Gamma)(x)\mathrm dx=\frac{\gamma}2+\sum_{k=2}^\infty\frac{\zeta(k)}{k(k+1)}$$


$$\int_0^1(\ln\Gamma)(x)\mathrm dx=\frac12\sum_{k=1}^\infty\frac{\zeta(2k)}{k(k+1)}$$


2 Answers 2


$$\frac{1}{2}\int_{0}^{1}\log \sin (\pi z)dz = \int_{0}^{1/2}\log \sin (\pi z)dz$$ $$\int_{0}^{1/2}\log \sin (\pi z)dz = \int_{0}^{1/2}\log \cos (\pi z)dz$$ Therefore, $$2\int_{0}^{1/2}\log \sin (\pi z)dz = \int_{0}^{1/2}\log \frac{\sin (2\pi z)}{2}dz = \int_{0}^{1/2}\log \sin (2\pi z)dz - \frac{1}{2}\ln 2$$ Substitute $2z = x$ in the first term on RHS to get, $$2\int_{0}^{1/2}\log \sin (\pi z)dz = \int_{0}^{1/2}\log \sin (\pi x)dx - \frac{1}{2}\ln 2$$ Thus, $$\frac{1}{2}\int_{0}^{1}\log \sin (\pi z)dz = - \frac{1}{2}\ln 2$$


The given integral equals

$$ \frac{1}{2}\int_{0}^{1}\log\left[\Gamma(x)\,\Gamma(1-x)\right]\,dx = \frac{1}{2}\int_{0}^{1}\log\frac{\pi}{\sin(\pi z)}\,dz \tag{1}$$ that is $$ \frac{1}{2}\log\pi -\frac{1}{2}\int_{0}^{1}\log\sin(\pi z)\,dz. \tag{2}$$ By a well-known symmetry argument$^{(*)}$ we have $\int_{0}^{1}\log\sin(\pi z)\,dz = -\log 2$ and the claim readily follows. We may reach the same conclusion by recalling that $$ \prod_{k=1}^{n-1}\sin\left(\frac{\pi k}{n}\right)=\frac{2n}{2^n}\tag{3} $$ taking logarithms and interpreting the LHS as a Riemann sum.
$(*)$ Let $I=\int_{0}^{1}\log\sin(\pi z)\,dz$. We have $I=2\int_{0}^{1/2}\log\sin(\pi z)\,dz$ but also $$ I = 2\int_{0}^{1/2}\log\sin(2\pi z)\,dz = \log(2)+2\int_{0}^{1/2}\log\sin(\pi z)\,dz+2\int_{0}^{1/2}\log\cos(\pi z)\,dz \tag{4}$$ from which $I=\log(2)+I+I$.

  • $\begingroup$ thank you but I dont know the argument that shows the identity related to $-\ln 2$. And the second identity is also unknown to me. $\endgroup$
    – Masacroso
    Commented Apr 10, 2017 at 8:53
  • $\begingroup$ @Masacroso: symmetry argument now explained. $(3)$ follows from standard manipulations on De Moivre's identity $\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$. $\endgroup$ Commented Apr 10, 2017 at 8:56
  • $\begingroup$ @Masacroso: out of curiosity, how did you prove the duplication or multiplication formula for the $\Gamma$ function without $(3)$? $\endgroup$ Commented Apr 10, 2017 at 8:58
  • $\begingroup$ starting from here $$\frac1{\Gamma(z)}=ze^{\gamma z}\prod_{k=1}^\infty\left(1+\frac{z}k\right)e^{-z/k}$$ that comes from $$\gamma_n(z)=\int_0^nt^{z-1}\left(1-\frac{t}n\right)^n\mathrm dt=\frac{n!n^z}{z(z+1)\cdots(z+n)}$$ and the book showed that $$\lim \gamma_n(z)=\Gamma(z)$$ $\endgroup$
    – Masacroso
    Commented Apr 10, 2017 at 9:02
  • $\begingroup$ @Masacroso: all right, thank you for the explanation. $\endgroup$ Commented Apr 10, 2017 at 9:05

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