Evaluate: $$I=\int_0^{\frac{π}{2}}\tan (x)\ln (\sin x)\ln (\cos x)dx$$

My ideas is to use the Fourier series of log sin and log cos:

$$\ln (2\sin x)=-\sum_{k=1}^{\infty}\frac{\cos (2kx)}{k}$$ $$\ln (2\cos x)=-\sum_{k=1}^{\infty}\frac{(-1)^{k}\cos (2kx)}{k}$$

But my problem is that I find difficult integrals like:

$$\int\tan (x)\cos (2kx)dx$$

My another idea is:

Use the substation : $y=\tan x$ then $dx=\frac{dy}{1+y^2}$

Then where $x=0 \Rightarrow y=0$ and for $x=\frac{π}{2} \Rightarrow y=\infty$


$$I=\frac{1}{2}\int_0^{\infty}\frac{y\ln \left(\frac{y}{\sqrt{1+y^2}}\right)\ln (1+y^2)}{1+y^2}dy$$

But now I don't know how to complete.

  • $\begingroup$ Thanks sir , but what happene I have $\tan x$ , $m,n>0$ $\endgroup$ – Tomas Houbaze Jul 7 at 7:27

$I=\int_0^{\frac{π}{2}}\tan (x)\ln (\sin x)\ln (\cos x)dx$

$=-\int_0^{-\infty}t\ln (\sqrt{1-e^{2t}})dt$

(substituting $\ln(\cos{x}) = t$, so $\tan{x} dx = -dt$, and $\sin{x} = \sqrt{1-\cos^2{x}} = \sqrt{1-e^{2t}}$)

$=-\int_0^{\infty}u\ln (\sqrt{1-e^{-2u}})du$ (Substituting $t = -u$)

$=-\frac{1}{2} \int_0^{\infty}u\ln (1-e^{-2u})du$

$=-\frac{1}{2}\int_0^{\infty}u\sum_{n=1}^{\infty} (-e^{-2nu}/n)du$

$=\int_0^{\infty}\sum_{n=1}^{\infty} (\frac{(2nu)e^{-2nu}}{8n^3})(2ndu)$

$=\sum_{n=1}^{\infty} \frac{1}{8n^3}\int_0^{\infty} ke^{-k} dk$ (Subtituting $2nu = k$, and $\int_0^{\infty} ke^{-k} dk =1 $)


  • $\begingroup$ Thank you very much sir @Archisman Panigrahi , sir if we have $\tan^{2} x$ or $\tan^{3} x$ not $\tan x$ then how I evaluate this ? $\endgroup$ – Tomas Houbaze Jul 7 at 8:07
  • $\begingroup$ Hi Thomas, please don't call me sir. I don't know how to do those integrals but I can think about it. Why don't you post it in this site? I am sure someone will be able to answer the question. $\endgroup$ – Archisman Panigrahi Jul 7 at 8:11
  • $\begingroup$ Mathematica says it does not converge for $\tan^3{x}$ but it does for $\tan^2{x}$ $\endgroup$ – Archisman Panigrahi Jul 7 at 8:15
  • $\begingroup$ Thanke you @Archisman Panigrahi $\endgroup$ – Tomas Houbaze Jul 7 at 8:21

$$\int_0^\frac{\pi}{2} \frac{\ln(\sin x)\ln(\cos x)}{\cos x}\sin xdx\overset{\cos x=t}=\int_0^1 \frac{\ln(\sqrt{1-t^2})\ln t}{t}dt\overset{t^2=x}=\frac18 \int_0^1 \frac{\ln(1- x)\ln x}{x}dx$$ $$=\frac18\int_0^1 \left(-\sum_{n=1}^\infty \frac{x^n}{n}\right)\frac{\ln x}{x}dx=-\frac18 \sum_{n=1}^\infty\frac{1}{n}\int_0^1 x^{n-1}\ln xdx=\frac18\sum_{n=1}^\infty \frac{1}{n^3}=\frac{\zeta(3)}{8}$$

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    $\begingroup$ Thank you very much sir @Zacky $\endgroup$ – Tomas Houbaze Jul 7 at 8:09
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    $\begingroup$ easy-peasy-integral-squeezy $\endgroup$ – clathratus Aug 2 at 2:55

We can actually use the Beta Function with this one. Here we will address your integral: \begin{equation} I = \int_0^\frac{\pi}{2} \tan(x) \ln(\sin(x))\ln(\cos(x))\:dx \end{equation} We note that: \begin{equation} \tan(x) = \sin(x)\cos^{-1}(x),\:\:\lim_{a \rightarrow 0^+} \frac{d}{da} \sin^a(x) = \ln(\sin(x)),\:\:\lim_{b \rightarrow 0^+} \frac{d}{db} \cos^b(x) = \ln(\cos(x)) \end{equation} Thus $I$ becomes: \begin{equation} I = \int_0^\frac{\pi}{2} \sin(x)\cos^{-1}(x) \cdot \lim_{a \rightarrow 0^+} \frac{d}{da} \sin^a(x) \cdot \lim_{b \rightarrow 0^+} \frac{d}{db} \cos^b(x)\:dx \end{equation} By the Dominated Convergence Theorem and Leibniz's Integral Rule: \begin{align} I &= \lim_{a \rightarrow 0^+} \frac{d}{da} \lim_{b \rightarrow 0^+} \frac{d}{db} \int_0^\frac{\pi}{2} \sin(x)\cos^{-1}(x) \sin^a(x) \cos^b(x)\:dx \nonumber \\ &= \lim_{(a,b)\rightarrow (0,0)^+} \frac{\partial^2}{\partial a \partial b} \int_0^\frac{\pi}{2} \sin^{a + 1}(x)\cos^{b - 1}(x)\:dx \nonumber \\ &= \lim_{(a,b)\rightarrow (0,0)^+} \frac{\partial^2}{\partial a \partial b} \left[\frac{1}{2}B\left(\frac{a + 2}{2}, \frac{b}{2} \right) \right] \end{align} Where $B(\cdot, \cdot)$ is the Beta Function

From here convert to the Gamma Representation, apply the derivatives and the evaluate at $(a,b) = (0,0)$ and you're done!

EDIT - This doesn't work. The Beta Function is undefined for $b = 0$.

  • $\begingroup$ Can you please explain how you get $\begin{equation} \lim_{a \rightarrow 0^+} \frac{d}{da} \sin^a(x) = \ln(\sin(x)),\:\:\lim_{b \rightarrow 0^+} \frac{d}{db} \cos^b(x) = \ln(\cos(x)) \end{equation}$ $\endgroup$ – Archisman Panigrahi Jul 7 at 12:16
  • $\begingroup$ $$ \frac{d}{da} \left[f(x)^a\right] = \ln(f(x))f(x)^a$$ Then evaluate at $a = 0$. $\endgroup$ – user679268 Jul 7 at 12:24
  • $\begingroup$ What is the relation between detectives of the gamma function and the zeta function? $\endgroup$ – Archisman Panigrahi Jul 7 at 14:34
  • $\begingroup$ Well it seems: $$ \frac{\zeta(3)}{8}=\lim_{(a,b)\rightarrow (0,0)^+} \frac{\partial^2}{\partial a \partial b} \left[\frac{1}{2}B\left(\frac{a + 2}{2}, \frac{b}{2} \right) \right] $$ $\endgroup$ – user679268 Jul 7 at 15:12
  • $\begingroup$ @ArchismanPanigrahi - I just evaluated it out and it's undefined at $b = 0$. I must have made a mistake somewhere in my work. $\endgroup$ – user679268 Jul 8 at 2:58

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