Solve $$I = \int_0^{\pi/2} \frac{\big(\log(\sin x)\big)^2\big(\log(\cos x)\big)^3}{\sin^2x\cos^2x}dx$$
Hello everybody! This is an integral that it has been bugging me for a while. I hope you guys can help me.
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2Welcome to MSE. Could you tell us what you have tried so far? – Gilles Bonnet Feb 13 '15 at 10:00
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In principle you you can solve this by considering the function
$$ J(a,b)=\int_{0}^{\pi/2}\sin^{a-2}(x)\cos^{b-2}(x)dx $$
Then your integral is given by $$ I=\partial^2_a\partial^3_b J(a,b)\big|_{a=b=0} $$
The integral $J(a,b)$ is a well known representation of Euler's Beta function so the result is
$$ I=\lim_{a,b\rightarrow0}\partial^2_a\partial^3_b\left[\frac{\Gamma[1/2 (a-1)] \Gamma[1/2 ( b-1)]}{2 \Gamma[ 1/2 (-2 + a + b)]}\right] $$
I leave the calculation of this limit to you because it is getting very exhausting. But the method should work, i tested it for some simpler cases.
tired
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5(+1) According to Mathematica, the value of the limit is given by: $$ -\frac{\pi ^3}{4}+\frac{\pi ^5}{96}+3 \pi \text{Log}[4]+\frac{1}{8} \pi ^3 \text{Log}[4]-3 \pi \text{Log}[4]^2-\frac{1}{16} \pi ^3 \text{Log}[4]^2+\frac{5}{4} \pi \text{Log}[4]^3-\frac{5}{16} \pi \text{Log}[4]^4-\frac{3}{4} \pi \text{Zeta}[3]+\frac{3}{4} \pi \text{Log}[4] \text{Zeta}[3]$$ – Jack D'Aurizio Feb 13 '15 at 10:59
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