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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?

Henry Lee
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3 Answers3

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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.

Jack D'Aurizio
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  • 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 – Henry Lee Sep 15 '18 at 22:21
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    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: $$B(a,b)=\int_{0}^{1}x^{a-1}(1-x)^{b-1},dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}.$$ – Jack D'Aurizio Sep 15 '18 at 22:25
  • can you check my edit to the question where I try this method – Henry Lee Sep 15 '18 at 22:39
  • 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? – Henry Lee Sep 16 '18 at 11:02
  • @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. – Jack D'Aurizio Sep 16 '18 at 16:40
  • so is this just from $\frac{d}{dx}\ln(f(x))=\frac{f'(x)}{f(x)}$? – Henry Lee Sep 16 '18 at 16:53
  • @HenryLee: exactly. Luckily $\Gamma$ has no complex zero. – Jack D'Aurizio Sep 16 '18 at 16:54
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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}

Ahmad Bazzi
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In general,

$$ \int_0^\frac{\pi}{2}\frac{\ln^m(\sin(x))\ln(\cos(x))}{\tan(x)}dx =\frac{(-1)^{m+1}}{2^{m+2}} \Gamma{(m+1)}\zeta(m+2) $$

Let $\sin x=e^{-t}$, then $$ \begin{aligned} \int_0^\frac{\pi}{2}\frac{\ln^m(\sin(x))\ln(\cos(x))}{\tan(x)}dx & =\frac{1}{2} \int_0^{\infty} (-t)^m \ln \left(1-e^{-2 t}\right) d t \\ & =\frac{(-1)^{m+1}}{2} \int_0^{\infty} t^m \sum_{n=1}^{\infty} \frac{e^{-2 n t}}{n} d t \\ & =\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n} \int_0^{\infty} t^m e^{-2 n t} d t \\ & = \frac{(-1)^{m+1}}{2} \sum_{n=1}^{\infty} \frac{1}{n} \frac{\Gamma(m+1)}{(2 n)^{m+1}} \\ & = \frac{(-1)^{m+1}}{2^{m+2}} \Gamma{(m+1)}\sum_{n=1}^{\infty} \frac{1}{n^{m+2}} \\ & = \frac{(-1)^{m+1}}{2^{m+2}} \Gamma{(m+1)}\zeta(m+2) \end{aligned} $$

Lai
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