\(\int_0^{\pi/2} \frac{\log(\cos x)}{\tan x }dx\)

Calculus (Integrals, Series)
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Grigorios Kostakos
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\(\int_0^{\pi/2} \frac{\log(\cos x)}{\tan x }dx\)

#1

Post by Grigorios Kostakos »

Evaluate
\[\displaystyle\int_{0}^{\frac{\pi}{2}} \frac{\log(\cos{x})}{\tan{x}}\, {\rm d}x\,.\]
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Tolaso J Kos
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Re: \(\int_0^{\pi/2} \frac{\log(\cos x)}{\tan x }dx\)

#2

Post by Tolaso J Kos »

Grigorios Kostakos wrote:Evaluate
\[\displaystyle\int_{0}^{\frac{\pi}{2}} \frac{\log(\cos{x})}{\tan{x}}\, {\rm d}x\,.\]
Hello Grigorios,

\begin{align*}
\int_{0}^{\pi/2} \frac{\log \cos x}{\tan x}\, {\rm d}x &=\cancelto{0}{\left [ \log \sin x \log \cos x \right ]_0^{\pi/2}} + \int_{0}^{\pi/2} \tan x \log \sin x \, {\rm d}x \\
&=\int_{0}^{\pi/2} \frac{\sin x}{\cos x} \log \sqrt{1-\cos^2 x}\, {\rm d}x \\
&\!\!\!\!\overset{u=\cos x}{=\! =\! =\! =\!} \frac{1}{2} \int_{0}^{1} \frac{\log \left ( 1-u^2 \right )}{u}\, {\rm d}u\\
&=-\frac{1}{2} \int_{0}^{1} \frac{1}{u} \sum_{n=1}^{\infty} \frac{u^{2n}}{n} \, {\rm d}u \\
&= -\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n}\int_{0}^{1}u^{2n-1} \, {\rm d}u \\
&=- \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^2} \\
&=- \frac{\pi^2}{24}
\end{align*}
Imagination is much more important than knowledge.
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Re: \(\int_0^{\pi/2} \frac{\log(\cos x)}{\tan x }dx\)

#3

Post by Tolaso J Kos »

A second not quite different approach for the initial integral. It is in the same spirit as the previous one.

\begin{align*}
\int^{\pi/2}_0\tan x\log \sin x{\rm d}x &=\int_{0}^1\frac{\ln\left(\sqrt{1-t^2}\right)}t\;{\rm d}t \tag{$t=\cos{x}$} \\
&=\int_{0}^1\frac{y\log y}{1-y^2}\;{\rm d}y\tag{$y=\sqrt{1-t^2}$} \\
&=\int_{0}^1y\log y\sum_{n=0}^{\infty}y^{2n}\; {\rm d}y \\
&=\sum_{n=0}^{\infty}\int_{0}^1y^{2n+1}\log y\, {\rm d}y \\
&=-\sum_{n=0}^{\infty}\frac1{(2n+2)^2} \\
&=-\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2} \\
&=-\frac{\pi^2}{24}
\end{align*}

It holds that:

$$\int_0^1 x^m \log^n x \, {\rm d}x = \frac{(-1)^n n!}{(m+1)^{n+1}}$$
Imagination is much more important than knowledge.
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