Show that \[ \int_{0}^{1} \log (\sin \pi x) d x=-\log 2 .\]
Hint: use the following contour.
Solution
1. \(\int_0^1 \log(\sin \pi x)\,dx = \frac{1}{\pi}\int_0^{\pi} \log(\sin t)\,dt\).
Substitute \(t = \pi x\).
2. \(\int_0^{\pi} \log(\sin t)\,dt = 2\int_0^{\pi/2} \log(\sin t)\,dt\).
The substitution \(t\mapsto\pi-t\) maps \([\pi/2,\pi]\) onto \([0,\pi/2]\) and \(\sin(\pi-t)=\sin t\).
3. \(\int_0^{\pi/2} \log(\sin t)\,dt = -\frac{\pi}{2}\log 2\).
2. Hence \(J = -\frac{\pi}{2}\log 2\).
4. \(\int_0^{\pi} \log(\sin t)\,dt = -\pi \log 2\).
Steps 2 and 3 give \(2 \cdot (-\frac{\pi}{2}\log 2)\).
5. \(\int_0^1 \log(\sin \pi x)\,dx = \boxed{-\log 2}\).
Steps 1 and 4 give \(\frac{1}{\pi}(-\pi \log 2)\).
6. Q.E.D.
Step 5 is the required identity.