how do I justify (which I sure is correct since I can evaluat this integral)?

$$\int_0^{\pi/2}\log\sin x\,\mathrm dx+\int_0^{\pi/2}\log\cos x\mathrm dx=\int_0^{\pi/2}(\log\cos x+\log\sin x)\mathrm dx $$

The problem is that I need to provied a proof that$$\int_0^{\pi/2}\log\sin x\,\mathrm dx>-\infty $$

  • $\begingroup$ You don't want to write $< \infty.$ You want to say it converges. $\endgroup$
    – zhw.
    Apr 14, 2017 at 19:35

2 Answers 2


Since $\lim_{x\to 0}(\sin x)/x=1$, we can take $A\in (0, \pi /2]$ such that $x\in (0,A]\implies \sin x>x/2.$ For any $B\in (0,A)$ and any $x\in [B,A]$ we have $0>\log \sin x >\log (x/2)$ so $$0>\int_B^A \log \sin x \;dx>\int^A_B\log (x/2)\;dx=$$ $$=(A\log A-A)-(B\log B-B)-(A-B)\log 2.$$ Now $\lim_{B\to 0^+} B\log B=0.$ So, letting $B\to 0^+$, we have $$0>\int_0^A \log \sin x\;dx\geq \int_0^A\log (x/2)\;dx=$$ $$=A\log A -A-A\log 2>-\infty.$$ And $\log \sin x$ is bounded on $[A,\pi /2].$

  • $\begingroup$ This is similar to the Comparison Test for series. Here, we look for a function $ f$ such that $\int_0^{\pi /2}|f(x)|dx<\infty$ and such that $|\log \sin x |<|f(x)|$ for all sufficiently small $x>0.$ $\endgroup$ Apr 14, 2017 at 20:36
  • $\begingroup$ thank u very much for your help $\endgroup$ Apr 14, 2017 at 20:46

Let $0\le x\le\frac{\pi}{2}$. By the Mean Value Theorem there exists $x'$, so that $\frac{sin(x)-sin(0)}{x-0}=\frac{d}{dx}(sin(x'))=cos(x')$ with $0\le x'\le x$. Then $$\begin{gather}\frac{sin(x)}{x}=cos(x')\Leftrightarrow sin(x)=cos(x')x\Leftrightarrow sin^2(x)=cos^2(x')x^2\Leftrightarrow\\ sin^2(x)=(1-sin^2(x'))x^2>(1-sin^2(x))x^2\end{gather}$$

The last inequality holds since $0\le x'\le x\le\frac{\pi}{2}$ implies that $sin^2(x')<sin^2(x)$, since $f(x)=sin(x)$ is strictly increasing on $[0;\frac{\pi}{2}]$. Now $$\begin{gather}sin^2(x)>(1-sin^2(x))x^2\Leftrightarrow sin^2(x)(1+x^2)>x^2\Leftrightarrow sin^2(x)>\frac{x^2}{1+x^2}\Leftrightarrow\\ sin(x)>\sqrt{\frac{x^2}{1+x^2}}=\frac{x}{\sqrt{1+x^2}}\end{gather}$$

We deduce that $\log\sin(x)>\log{\frac{x}{\sqrt{1+x^2}}}=\log x-\log\sqrt{1+x^2}=\log x-\frac{1}{2}\log(1+x^2)$ on the interval $[0;\frac{\pi}{2}]$ due to the $\log$ being a monotonically increasing function.

Therefore $$\int_0^\frac{\pi}{2}\log\sin xdx>\int_0^\frac{\pi}{2}\log x-\frac{1}{2}\log(1+x^2)dx=\int_0^\frac{\pi}{2}\log xdx-\frac{1}{2}\int_0^\frac{\pi}{2}\log(1+x^2)dx=[x(\log x-1)]_0^\frac{\pi}{2}-\frac{1}{2}[x(\ln(x^2+1)-2)+2\arctan(x)]_0^\frac{\pi}{2}\approx-0.86+0-\frac{1}{2}((-1.188+2.008)-(0+0))\approx-1.27>-\infty$$

  • $\begingroup$ I think that as in series linearity holds iff $\int_0^{\pi/2}\log\sin x\,\mathrm dx<\infty$ $\endgroup$ Apr 14, 2017 at 19:14
  • $\begingroup$ I think linearity also applies when the integral diverges, but I wouldn't want to swear on that. To be sure, prove its converge: Consider $\log x<x$ for $x>0$ and $\sin x<x$ for $x\ge0$ and putting both together $\log\sin x<x$ for $x>0$ and therefore $\int_0^{\frac{\pi}{2}}\log\sin x dx<\int_0^{\frac{\pi}{2}}x dx=\frac{\pi^2}{8}$. $\endgroup$
    – Thorgott
    Apr 14, 2017 at 19:31
  • $\begingroup$ I didn't been completly accurate, the problem is to prove that $\int_0^{\pi/2}\log\sin x\,\mathrm dx>-\infty$ $\endgroup$ Apr 14, 2017 at 19:38

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