Show that the improper Riemann integral \[ I=\int_0^\pi\log(\sin x)\,dx \] converges, and evaluate \(I\).
Solution
1. The improper integral converges absolutely.
For \(0<x\leq\pi/2\), concavity of \(\sin x\) on \([0,\pi]\) gives the chord bound \[ \sin x\geq\frac{2x}{\pi}. \] Since \(0<\sin x\leq1\) on this interval, \[ 0 \leq -\log(\sin x) \leq -\log\left(\frac{2x}{\pi}\right). \] The majorant is integrable at \(0\), because \[ \int_0^{\pi/2} -\log\left(\frac{2x}{\pi}\right)\,dx <\infty. \] Hence \[ \int_0^{\pi/2}\abs{\log(\sin x)}\,dx<\infty. \]
For \(\pi/2\leq x<\pi\), put \(u=\pi-x\). Then \(\sin x=\sin u\), so the same estimate applies at \(\pi\). Thus \[ \int_0^\pi\abs{\log(\sin x)}\,dx<\infty. \] In particular, the stated improper Riemann integral converges.
2. If\[ J\coloneqq\int_0^{\pi/2}\log(\sin x)\,dx, \]then\[ I=2J. \]
The substitution \(u=\pi-x\) gives \[ \int_{\pi/2}^{\pi}\log(\sin x)\,dx = \int_0^{\pi/2}\log(\sin u)\,du =J. \] The substitutions are legitimate by step 1.
3. One also has\[ J = \int_0^{\pi/2}\log(\cos x)\,dx. \]
Use the substitution \(u=\pi/2-x\) and the identity \[ \sin(\pi/2-u)=\cos u. \]
4. The number \(J\) satisfies\[ 2J = -\frac{\pi}{2}\log2+J. \]
By step 3, \[ \begin{aligned} 2J &= \int_0^{\pi/2} \log(\sin x\cos x)\,dx\\ &= \int_0^{\pi/2} \log\left(\frac{\sin(2x)}2\right)\,dx\\ &= -\frac{\pi}{2}\log2 + \int_0^{\pi/2}\log(\sin(2x))\,dx. \end{aligned} \] With \(u=2x\) and step 2, \[ \int_0^{\pi/2}\log(\sin(2x))\,dx = \frac12\int_0^\pi\log(\sin u)\,du = \frac12 I =J. \] Substitution gives the claimed equation.
5. Therefore\[ \boxed{ I = -\pi\log2 }. \]
Step 4 gives \[ J=-\frac{\pi}{2}\log2. \] Apply \(I=2J\) from step 2.
6. Q.E.D.
Step 1 proves convergence and step 5 evaluates the integral.