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Let $I=\int_0^{\pi/2} \log(\sin x)dx$. Then

  1. I diverges at $x=0$.

  2. I converges and is equal to $-\pi\log 2$.

  3. I converges and is equal to $-\pi/2 \log 2$.

  4. I diverges at $x=\pi/4$.

I found that value of integral is zero. and also that $\log(\sin x)$ is not a decreasing function, so we can not use integral test here. I am not able to think of applying any other series convergence test here. Please, help me out.

Robert Z
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Parul
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1 Answers1

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The answer is the third one.

We have that $$I=\int_0^{\pi/2} \log(\sin x)dx=\int_0^{\pi/2} \log(\sin(\pi/2-x)dx=\int_0^{\pi/2} \log(\cos x)dx.$$ Moreover $$\int_0^{\pi} \log(\sin x)dx=I+\int_{\pi/2}^{\pi} \log(\sin x)dx \\= I+\int_0^{\pi/2} \log(\sin (\pi-x))dx\\= I+\int_0^{\pi/2} \log(\sin x)dx=2I.$$ Hence $$2I=\int_0^{\pi/2} \log(\sin x)dx+\int_0^{\pi/2} \log(\cos x)dx= \int_0^{\pi/2} \log(\sin x\cos x)dx\\=\int_0^{\pi/2} \log\left(\frac{\sin 2x}{2}\right)dx=\int_0^{\pi/2} \log(\sin 2x)dx-\int_0^{\pi/2} \log(2)dx\\ =\frac{1}{2}\int_0^{\pi} \log(\sin t)dt-\frac{\pi}{2}\log(2) =\frac{2I}{2}-\frac{\pi}{2}\log(2)$$ which implies that $I=-\frac{\pi}{2}\log(2)$.

Robert Z
  • 145,942