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I've seen many proof of $$\int_0^{\pi/2} \log\sin x\,\mathrm dx= -\dfrac{\pi}{2} \log 2$$ where result was obtained through algebraic manipulations... But to do this, we must to prove that the integral converges.

My question is : how to prove that $\int_0^{\pi/2} \log\sin x\,\mathrm dx$ converges?

Thanks for answers.

SuperFoxy
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    Using the relation $\frac{2}{\pi}x \leq \sin x \leq x$ for $x \in [0, \frac{\pi}{2}]$, we check that $\log \sin x = \log x + O(1)$. Then it is straightforward to prove that $\log x$ is integrable on $[0, \frac{\pi}{2}]$ (both in improper Riemannn sense and in Lebesgue sense). – Sangchul Lee Oct 26 '16 at 10:28

4 Answers4

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The integrand is negative so you can use the comparison test (if you prefer to work with positive integrands, rewrite it as $-\int_0^{\frac{\pi}{2}} (-\log \sin x) \, dx$). Thus, we can compare your integral with the integral $\int_0^1 \log(x) \, dx$ which can be evaluated explicitly:

$$ \int_0^1 \log(x) \, dx = \lim_{\delta \to 0} \int_{\delta}^1 \log(x) \, dx = \lim_{\delta \to 0} \left[ x \log(x) - x\right]^{x = 1}_{x = \delta} = -1 -\delta \log \delta + \delta \xrightarrow[\delta \to 0]{} -1$$

Since

$$ \lim_{x \to 0} \frac{\log \sin x}{\log x} = \lim_{x \to 0} \frac{\frac{\cos x}{\sin x}}{\frac{1}{x}} = \left( \lim_{x \to 0} \cos x \right) \left( \lim_{x \to 0} \frac{x}{\sin x} \right) = 1$$

the integrals must converge/diverge together, and in this case, they converge together.

levap
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For $0<x<1$,

$\log(\sin(x))=$

$\log(x)(1+\frac{1}{\log(x)}\log(\frac{\sin(x)}{x}))=$

$\log(x)(1+\epsilon(x))$

with $\lim_{x\to 0^+}\epsilon(x)=0$.

So, by equivalence criteria, near $0^+$, our integral is convergent since $\int_0log(x)dx $ is concergent.

To compute its value, let

$$S=\int_0^\frac{\pi}{2}\log(\sin(x))dx$$

And $$C=\int_0^\frac{\pi}{2}\log(\cos(x))dx$$

With the change $y=\frac{\pi}{2}-x$, we get $C=S$.

On the other hand

$$S+C=\int_0^\frac{\pi}{2}(\log(\sin(2x)-\log(2))dx$$

or, with $t=2x$

$$=\frac{1}{2}\int_0^\pi \log(\sin(t))dt-\frac{\pi}{2}\log(2)$$

$$\frac{1}{2}(S+\int_\frac{\pi}{2}^\pi \log(\sin(t))dt)-\frac{\pi}{2}\log(2)$$

and finally, with $u=\pi-t$, we get

$$\color{green}{S=C=-\frac{\pi}{2}\log(2)}$$

hamam_Abdallah
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Thanks to Sangchul Lee comment : on $[0, \pi/2]$ $$ \begin{split} x\geq \sin x \geq \frac{2}{\pi} x \Rightarrow \int_0^{\pi/2} \log x\,\mathrm dx = \frac{\pi}{2}\log(\pi/2)-\frac{\pi}{2} \geq \int_0^{\pi/2} \log \sin x\,\mathrm dx &\geq \frac{\pi}{2}\log(2/\pi)+ \int_0^{\pi/2} \log x\,\mathrm dx\\&= \frac{\pi}{2}\log(2/\pi) + \frac{\pi}{2}\log(\pi/2)-\frac{\pi}{2} \end{split}$$ The integral $\int_0^{\pi/2} \log \sin x\,\mathrm dx$ is bounded, thus converges.

SuperFoxy
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The problem is when $x\to 0^+$, as $\log (\sin x)\to-\infty$. Since $\log (\sin x)= \log x +\text o(x^2)$, you have: $$\int_0^{\pi/2}\log (\sin x)dx\sim \int_0^{\pi/2} \log x dx$$ and the last integral is convergent.

Sine of the Time
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