\(\int_{0}^{\frac{\pi}{2}}{\log^2(\tan{x})\,dx}\)

Calculus (Integrals, Series)
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Grigorios Kostakos
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\(\int_{0}^{\frac{\pi}{2}}{\log^2(\tan{x})\,dx}\)

#1

Post by Grigorios Kostakos »

Calculate the integral \[\displaystyle \int_{0}^{\frac{\pi}{2}}{\log^2(\tan{x})\,dx}\,.\]
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Tolaso J Kos
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Re: \(\int_{0}^{\frac{\pi}{2}}{\log^2(\tan{x})\,dx}\)

#2

Post by Tolaso J Kos »

Good evening!

We use the fact that $\displaystyle \int_0^{\pi/2} \ln^2 (\sin x) \, {\rm d}x = \frac{\pi^3}{24} + \frac{\pi \ln^2 2}{2}$ which was proven here. Now it is easy to see that:

$$\int_{0}^{\pi/2}\ln^2 \left ( \cos x \right )dx \overset{u=\pi/2-x}{=\! =\! =\! =\! =\! =\!}\int_{0}^{\pi/2}\ln^2 \left ( \sin x \right )dx=\frac{\pi^3}{24}+\frac{\pi}{2}\ln^2 2$$

We also see that the integrand function is well defined, coz \( \tan x>0, \,\,\,\,\, x\in \left ( 0, \pi/2 \right ) \) . Assuming that the integral does converge we are going to evaluate it.

\begin{align*}
\int_{0}^{\pi/2}\left ( \ln \tan x \right )^2 \, {\rm d}x &=\int_{0}^{\pi/2} \left ( \ln \frac{\sin x}{\cos x} \right )^2\, {\rm d}x \\
&= \int_{0}^{\pi/2}\left ( \ln \sin x -\ln \cos x \right )^2 \, {\rm d}x\\
&=\int_{0}^{\pi/2}\ln^2 \sin x \, {\rm d}x + \int_{0}^{\pi/2}\ln^2\cos x\, {\rm d}x - 2\int_{0}^{\pi/2}\ln \sin x \ln \cos x \, {\rm d}x \\
&= \frac{\pi^3}{12}+ \pi\ln^2 2 - 2\int_{0}^{\pi/2}\ln \sin x \ln \cos x \, {\rm d}x \quad \quad (\color{blue}\bigstar )
\end{align*}

Now, all we have to do is evaluate the integral: \( \displaystyle \int_{0}^{\pi/2}\ln \sin x \ln \cos x \, dx \) and we will have evaluated the original integral.
So here goes:

$$\begin{aligned}
\int_{0}^{\pi/2}\ln \sin x \ln \cos x \, dx &=2\int_{0}^{\pi/4}\ln \sin x \ln \cos x \, dx \\
&\overset{u=\ln \sin x}{=\! =\! =\! =\! =\!}\frac{1}{2}\int_{-\infty }^{0}\frac{xe^x\ln \left ( 1-e^{2x} \right )}{\sqrt{1-e^{2x}}}dx \\
&\overset{y=1-e^{2x}}{=\! =\! =\! =\! =\!}\frac{1}{8}\int_{0}^{1} \frac{\ln (1-x)\ln x}{\sqrt{x}\sqrt{1-x}}dx
\end{aligned}$$

What follows now is not my solution, but found on an old text calculus book and a hint.

We'll use the known idenity \( \displaystyle {\rm B}(s, w+1)=\frac{\Gamma (s)\Gamma (w)}{\Gamma (s+w+1)} \) and differentiate.
Thus

$$\frac{d}{ds}\frac{d}{dw}\int_{0}^{1}\left ( 1-u \right )^wu^{s-1}du=\frac{d}{ds}\frac{d}{dw} {\rm B}(s, w+1)=\frac{d}{ds}\frac{d}{dw}\frac{\Gamma (s)\Gamma (w)}{\Gamma (s+w+1)}=u^{s-1}(1-u)^w\ln u \ln (1-u) $$

Now,

$$\frac{d}{ds}\frac{d}{dw}\frac{\Gamma (s)\Gamma (w+1)}{\Gamma (s+w+1)}= \frac{\Gamma (s)\Gamma (w+1)}{\Gamma (s+w+1)}\left ( \psi (s)\psi (w+1)-\psi (w+1)\psi (w+s+1)+\psi (s+w+1)^2-\psi '(s+w+1) \right )$$


If we let \( \displaystyle s=\frac{1}{2}, \,\,\,\, w=-\frac{1}{2} \) we get the result \( \displaystyle 2\left ( \frac{\pi\ln^22}{2}-\frac{\pi^3}{48} \right )=\pi\ln^2 2-\frac{\pi^3}{24} \) .

Summing up, \( \displaystyle (\color{blue}\bigstar ) \) gives, after all calculations are done,

$$\int_{0}^{\pi/2}\ln^2 \tan x\, dx=\frac{\pi^3}{8}$$
Imagination is much more important than knowledge.
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Re: \(\int_{0}^{\frac{\pi}{2}}{\log^2(\tan{x})\,dx}\)

#3

Post by Grigorios Kostakos »

A 2nd solution:

\begin{align*}
I_2&=\displaystyle \int_{0}^{\frac{\pi}{2}}{\log^2(\tan{x})\,dx}\\
&\mathop{=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{t\,=\,\tan{x}}\\
{dx\,=\,\frac{1}{1+t^2}\,dt}\\
\end{subarray}}\,\int_{0}^{+\infty}{\frac{\log^2{t}}{1+t^2}\,dt}\\
&=\int_{0}^{1}{\frac{\log^2{t}}{1+t^2}\,dt}+\int_{1}^{+\infty}{\frac{\log^2{t}}{1+t^2}\,dt}\\
&\mathop{=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{s\,=\,\frac{1}{t}}\\
{-\frac{1}{s^2}\,ds\,=\,dt}\\
\end{subarray}}\,\int_{0}^{1}{\frac{\log^2{t}}{1+t^2}\,dt}-\int_{1}^{0}{\frac{\log^2{\tfrac{1}{s}}}{1+\frac{1}{s^2}}\,\frac{1}{s^2}\,ds}\\
&=\int_{0}^{1}{\frac{\log^2{t}}{1+t^2}\,dt}+\int_{0}^{1}{\frac{\log^2{s}}{1+s^2}\,ds}\\
&=2\,\int_{0}^{1}{\frac{\log^2{t}}{1+t^2}\,dt}\\
&\mathop{=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{t\,=\,{\mathrm{e}}^{-x}}\\
{dt\,=\,-{\mathrm{e}}^{-x}dx}\\
\end{subarray}}\,-2\,\int_{+\infty}^{0}{\frac{(-x)^2\,{\mathrm{e}}^{-x}}{1+{\mathrm{e}}^{-2x}}\,dx}\\
&=2\,\int_{0}^{+\infty}{x^2\frac{{\mathrm{e}}^{-x}}{1+{\mathrm{e}}^{-2x}}\,dx}\\
&=2\,\int_{0}^{+\infty}{x^2\,\biggl({\mathop{\sum}\limits_{n=1}^{+\infty}{(-1)^{n+1}\,\mathrm{e}}^{-(2n-1)x}}\biggr)\,dx}\\
&=2\,\mathop{\sum}\limits_{n=1}^{+\infty}\biggl({(-1)^{n+1}\int_{0}^{+\infty}{x^2\,\mathrm{e}}^{-(2n-1)x}\,dx}\biggr)\\
&\mathop{=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{t\,=\,(2n-1)x}\\
{\frac{1}{2n-1}\,dt\,=\,dx}\\
\end{subarray}}\,2\,\mathop{\sum}\limits_{n=1}^{+\infty}\biggl({(-1)^{n+1}\,\int_{0}^{+\infty}{\frac{t^2}{(2n-1)^3}\,{\mathrm{e}}^{-t}\,dt}}\biggr)\\
&=2\,\mathop{\sum}\limits_{n=1}^{+\infty}\biggl({\frac{(-1)^{n+1}}{(2n-1)^3}\int_{0}^{+\infty}{t^{3-1}\,\mathrm{e}}^{-t}\,dt}\biggr)\\
&=2\,\mathop{\sum}\limits_{n=1}^{+\infty}\biggl({\frac{(-1)^{n+1}}{(2n-1)^3}\,\Gamma(3)}\biggr)\\
&=4\,\mathop{\sum}\limits_{n=1}^{+\infty}{\frac{(-1)^{n+1}}{(2n-1)^3}}\\
&=4\,\frac{\pi^3}{32}\\
&=\frac{\pi^3}{8}\,.
\end{align*}

Note: With similar methods we can calculate \(I_{2n}=\displaystyle \int_{0}^{\frac{\pi}{2}}{\log^{2n}(\tan{x})\,dx}\,,\quad n\in{\mathbb{N}}\,.\)
We have also that \(I_{2n-1}=\displaystyle \int_{0}^{\frac{\pi}{2}}{\log^{2n-1}(\tan{x})\,dx}=0\,,\quad n\in{\mathbb{N}}\,.\)
Grigorios Kostakos
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