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I have the problem below: $$\int_0^\frac{\pi}{2}\frac{\ln(\sin(x))\ln(\cos(x))}{\tan(x)}dx$$ I have tried $u=\ln(\sin(x))$ so $dx=\tan(x)du$

so the integral becomes: $$\int_{-\infty}^0u\ln(\cos(x))du$$ but I cannot find a simple way of getting rid of this $\ln(\cos(x))$

I also tried using the substitution $v=x-\frac{\pi}{2}$ so the integral becomes: $$\int_{-\frac{\pi}{2}}^0\frac{\ln(\cos(v))\ln(\cos(v+\frac{\pi}{2}))}{\tan(v+\frac{\pi}{2})}dv$$ but this does not seems to lead anywhere useful

EDIT to follow up

$$B(\alpha,\beta)=\int_0^{\pi/2}\sin^{\alpha-1}(x)\cos^{\beta+1}(x)dx$$ $$\frac{\partial^2}{\partial_\alpha\partial_\beta}B(\alpha,\beta)=\int_0^{\pi/2}\sin^{\alpha-1}(x)\cos^{\beta+1}(x)\ln(\sin(x))\ln(\cos(x))dx$$ so I see that when $\alpha\to1$ and $\beta\to-1$ that this is the form that we want, so do we now have to take partial derivates of the non-integral form of the beta function then take the double integral for our chosen values?

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  • $\begingroup$ Then you take the explicit form of $B(\alpha,\beta)$, which is given by a ratio of $\Gamma$ functions, differentiate it twice and evaluate it at the right points. $\endgroup$ – Jack D'Aurizio Sep 15 '18 at 22:41
  • $\begingroup$ How do you take the partial derivatives of the Gamma function? $\endgroup$ – Henry Lee Sep 15 '18 at 22:42
  • $\begingroup$ With the chain rule. $\frac{d}{dx}\Gamma(x)=\Gamma(x)\psi(x)$. $\endgroup$ – Jack D'Aurizio Sep 15 '18 at 22:43
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\begin{equation} \int_0^\frac{\pi}{2}\frac{\ln(\sin(x))\ln(\cos(x))}{\tan(x)}dx = \int_0^\frac{\pi}{2} \cos x \frac{\ln(\sin(x))\ln(\sqrt{1 - \sin^2 x})}{\sin(x)}dx \end{equation} Take \begin{equation} u = \sin x \end{equation} so \begin{equation} du = \cos x dx \end{equation} You get \begin{equation} \int_0^1 \ln u \ln(\sqrt{1- u^2}) \frac{1}{u} du \end{equation} Take $u = e^v$, you get \begin{equation} \frac{1}{2} \int_{-\infty}^0 v \ln(1 - e^{2v}) dv \end{equation} Integration by parts will give you \begin{equation} \frac{1}{2} [\frac{1}{2}v^2 \ln(1 - e^{2v})]_{-\infty}^0 - \frac{1}{2} \int_{-\infty}^0 \frac{v^2}{1- e^{2v}} (-2e^{2v}) dv \end{equation} The term $ [\frac{1}{2}v^2 \ln(1 - e^{2v})]_{-\infty}^0$ is zero so \begin{equation} - \frac{1}{2} \int_{-\infty}^0 \frac{v^2}{1- e^{2v}} (-2e^{2v}) dv \end{equation} Take the following change of variable \begin{equation} k = -v \end{equation} So we get \begin{equation} \frac{1}{2} \int_0^{\infty} \frac{k^2}{e^{2k} - 1} dK \end{equation} Take now \begin{equation} n = 2k \end{equation} as change of variable, you get \begin{equation} \frac{1}{2} \int_0^{\infty} \frac{k^2}{e^{2k} - 1} dk = \frac{1}{16} \int_0^{\infty} \frac{n^2}{e^n - 1} \end{equation} Using the Riemann Zeta function
\begin{equation} \zeta(x) = \frac{1}{\Gamma(x)} \int_0^{\infty} \frac{u^{x-1}}{e^u - 1} du \end{equation} or \begin{equation} \zeta(x)\Gamma(x) = \int_0^{\infty} \frac{u^{x-1}}{e^u - 1} du \end{equation} Using the above integral for $u = n$ and $x = 3$ you will get that your integral evaluates to \begin{equation} \frac{1}{2} \int_0^{\infty} \frac{k^2}{e^{2k} - 1} dk = \frac{1}{16} \int_0^{\infty} \frac{n^2}{e^n - 1} = \frac{\overbrace{\Gamma(3)}^{2!}\zeta(3)}{16} = \frac{\zeta(3)}{8} \end{equation}

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    $\begingroup$ ah I see, that is a clever method. How did you get this transformation for the zeta function? $\endgroup$ – Henry Lee Sep 15 '18 at 22:23
  • $\begingroup$ which equation exactly ? $\endgroup$ – Ahmad Bazzi Sep 15 '18 at 22:29
  • $\begingroup$ stating the equation for $\zeta(x)\Gamma(x)$ is this a known formula? $\endgroup$ – Henry Lee Sep 15 '18 at 22:30
  • $\begingroup$ I have linked a reference above $\endgroup$ – Ahmad Bazzi Sep 15 '18 at 22:31
  • $\begingroup$ Thank you thats great $\endgroup$ – Henry Lee Sep 15 '18 at 22:32
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You may consider that $$ f(\alpha,\beta) = \int_{0}^{\pi/2}\sin^{\alpha-1}(x)\cos^{\beta+1}(x)\,dx = \frac{\Gamma\left(\frac{a}{2}\right)\Gamma\left(1+\frac{b}{2}\right)}{2\,\Gamma\left(1+\frac{a+b}{2}\right)}$$ by Euler's Beta function. By applying $\frac{\partial^2}{\partial\alpha\,\partial\beta}$ to both sides, then considering the limit as $\beta\to 0$ and $\alpha\to 0^+$, we have

$$ \int_{0}^{\pi/2}\frac{\log\sin(x)\log\cos(x)}{\tan(x)}\,dx = -\frac{1}{12}\psi''(1) = \color{red}{\frac{\zeta(3)}{8}}.$$ By enforcing the substitutions $x\mapsto\arctan(x)$ or $x\mapsto 2\arctan(x)$ in the original integral we get interesting identities for nasty (poly)logarithmic integrals.

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  • $\begingroup$ I have never used the Beta function but I think I understand what it is. what is this double-prime function that you state in the last line and how does the limit relate to the $\ln$ function $\endgroup$ – Henry Lee Sep 15 '18 at 22:21
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    $\begingroup$ Here $\psi$ does not stand for the Chebyshev function. The digamma function $\psi(x)$ is defined as $\frac{d}{dx}\log\Gamma(x)=\frac{\Gamma'(x)}{\Gamma(x)}$. About the Beta function, it might be a good moment for learning its definition and typical usage: $$B(a,b)=\int_{0}^{1}x^{a-1}(1-x)^{b-1}\,dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}.$$ $\endgroup$ – Jack D'Aurizio Sep 15 '18 at 22:25
  • $\begingroup$ can you check my edit to the question where I try this method $\endgroup$ – Henry Lee Sep 15 '18 at 22:39
  • $\begingroup$ I now have $I=\lim_{\beta\to 0}\lim_{\alpha\to 0}\frac{\partial^2}{\partial_\alpha\partial_\beta}\left[\frac{\Gamma(\frac{\alpha}{2})\Gamma(1+\frac{\beta}{2})}{2\Gamma(1+\frac{\alpha + \beta}{2})}\right]$ but the differential of this explicit form is very complex, is there an easier way of evaluating it? $\endgroup$ – Henry Lee Sep 16 '18 at 11:02
  • $\begingroup$ @HenryLee: not so complex if you exploit $\frac{d}{dx}f(x)=f(x)\cdot\frac{d}{dx}\log f(x)$. This is pretty practical every time a function is (implicitly) defined by a (infinite) product, Euler docet. $\endgroup$ – Jack D'Aurizio Sep 16 '18 at 16:40

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