Using the duplication formula for \(\Gamma(2z)\), prove that \[ \int_0^1\log\Gamma(t)\,dt =\log\sqrt{2\pi}. \]
Solution
Set \[ I=\int_0^1\log\Gamma(t)\,dt. \] Then \[ \boxed{I=\frac12\log(2\pi)=\log\sqrt{2\pi}}. \]
1. The integral defining \(I\) converges.
For \(t>0\), the Gamma function is positive, so \(\log\Gamma(t)\) is well-defined and real-valued. The only possible endpoint issue is at \(t=0\). From the functional equation \[ \Gamma(t+1)=t\Gamma(t) \] and \(\Gamma(1)=1\), we have \[ \Gamma(t)=\frac{\Gamma(t+1)}{t}. \] As \(t\downarrow0\), \(\Gamma(t+1)\to1\), so \[ \log\Gamma(t)=-\log t+O(1). \] Since \(-\log t\) is integrable at \(0\), the improper integral converges.
2. The duplication formula implies, for \(0<t<1/2\),\[ \log\Gamma(t)+\log\Gamma(t+1/2) =\log\Gamma(2t)+(1-2t)\log2+\frac12\log\pi. \]
The Legendre duplication formula is \[ 2\sqrt\pi\,\Gamma(2t) =2^{2t}\Gamma(t)\Gamma(t+1/2). \] Rearranging gives \[ \Gamma(t)\Gamma(t+1/2) =2^{1-2t}\sqrt\pi\,\Gamma(2t). \] All terms are positive for \(t>0\), so taking logarithms gives the displayed identity.
3. Integrating the identity in 2 over \(0<t<1/2\) gives\[ I=\frac12I+\frac14\log(2\pi). \]
The left-hand side integrates to \[ \begin{aligned} &\int_0^{1/2}\log\Gamma(t)\,dt +\int_0^{1/2}\log\Gamma(t+1/2)\,dt\\ &\qquad= \int_0^{1/2}\log\Gamma(t)\,dt +\int_{1/2}^1\log\Gamma(u)\,du =I. \end{aligned} \]
For the first term on the right-hand side of 2, put \(u=2t\): \[ \int_0^{1/2}\log\Gamma(2t)\,dt =\frac12\int_0^1\log\Gamma(u)\,du =\frac12I. \] Also, \[ \int_0^{1/2}(1-2t)\,dt=\frac14, \qquad \int_0^{1/2}\frac12\log\pi\,dt=\frac14\log\pi. \] Therefore \[ I =\frac12I+\frac14\log2+\frac14\log\pi =\frac12I+\frac14\log(2\pi). \]
4. Solving for \(I\) yields the claimed value.
From 3, \[ \frac12I=\frac14\log(2\pi), \] so \[ I=\frac12\log(2\pi)=\log\sqrt{2\pi}. \]