I have to find the value of $$\int_{0}^{\frac{\pi}{2}}\log(\cos(x))\log(\sin(x))dx$$
in terms of $\pi$ and $\log(2)$. Any hint?
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3A parenthesis is missing in your integral, so it is not clear what the integrating function should be. Is this question supposed to be a challenge? – b00n heT May 13 '16 at 13:27
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1Yes it should be a challenge and the question is edited – Shivam Jadhav May 13 '16 at 15:06
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2It is an interesting problem, but you should show your attempts to solve it. – Jack D'Aurizio May 13 '16 at 15:31
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By Feynman's trick/differentiation under the integral sign and Euler's beta function: $$I=\int_{0}^{\pi/2}\log(\sin x)\log(\cos x)\,dx = \frac{\partial^2}{\partial a\,\partial b}\left.\frac{\Gamma\left(\frac{1+a}{2}\right)\Gamma\left(\frac{1+b}{2}\right)}{2\,\Gamma(\frac{a+b+2}{2})}\right|_{(a,b)=(0,0)}\tag{1}$$ Since $\Gamma'(x) = \psi(x)\,\Gamma(x)$, we have: $$ I = \lim_{(a,b)\to (0,0)}\frac{\Gamma\left(\frac{1+a}{2}\right)\Gamma\left(\frac{1+b}{2}\right)}{8\,\Gamma(\frac{a+b+2}{2})}\left[\psi\left(\frac{1+a}{2}\right)\psi\left(\frac{1+b}{2}\right)+\ldots-\psi'\left(\frac{a+b+2}{2}\right)\right]\tag{2}$$ and by exploiting: $$ \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi},\qquad \psi(1)=-\gamma,\qquad \psi'(1)=\zeta(2)=\frac{\pi^2}{6} \tag{3}$$ we get:
$$ I = -\frac{\pi^3}{48}-\frac{\pi\log 2}{4}\left[\gamma+\psi\left(\frac{1}{2}\right)\right] =\color{red}{\frac{\pi}{2}\log^2 2-\frac{\pi^3}{48}}.\tag{4}$$
Jack D'Aurizio
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Just curious did you look up the trick or did you just know how to do it? – Wolfy May 13 '16 at 15:36
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@Wolfy: it is a well-known trick to deal with logarithmic integrals, I used it many times here on MSE. – Jack D'Aurizio May 13 '16 at 15:37