Over the past week I've come across several 'Natural Logarithm Integrals': $$\int_0^{\pi/2}\ln(\sin(x))\tan(x)dx, \int_0^{\pi/2}\ln(\sin(x))\ln(\tan(x))dx$$ and so on so forth. This lead to me creating a question:
Evaluate $$I=\int_0^{\pi/2} \ln^3(\cot(x))\ln^3(\sec(x)) dx$$
To solve this, I tried the following approach:$$$$ Consider $$\beta\bigg (\dfrac{1-a}{2},\dfrac{a-b+1}{2}\bigg ) =2\int_0^{\pi/2} \sin^{-a}(x)\cos^{a}(x)\cos^{-b}(x)dx$$ $$$$ $$=\dfrac{\Gamma \bigg (\dfrac{1-a}{2}\bigg )\Gamma \bigg (\dfrac{a-b+1}{2}\bigg )}{\Gamma \bigg (\dfrac{2-b}{2}\bigg )}$$ $$$$ $$\Rightarrow \dfrac{1}{2} \dfrac{\partial^6}{\partial a^3 \partial b^3}\beta\bigg (\dfrac{1-a}{2},\dfrac{a-b+1}{2}\bigg ) \bigg|_{a=0 , b=0} = I$$ $$$$ $$=\dfrac{1}{2} \dfrac{\partial^6}{\partial a^3 \partial b^3}\dfrac{\Gamma \bigg (\dfrac{1-a}{2}\bigg )\Gamma \bigg (\dfrac{a-b+1}{2}\bigg )}{\Gamma \bigg (\dfrac{2-b}{2}\bigg )} \bigg |_{a=0 , b=0}$$ $$$$ Unfortunately, this method would become really, really messy and tedious. I was thus hoping for a better, neater method to find the closed form of the Integral. $$$$ I would be extremely grateful if somebody could please help me solve this problem. Many, many thanks in advance for your time and consideration!
After many calculations, but without difficulties Ifind $$I=-(\frac{15}{64}{\pi^5}\ln^22+\frac{63}{16}{\pi^3}\zeta{(3)}\ln2+\frac{279}{8}{\pi}\zeta{(5)}\ln2+\frac{441}{32}{\pi}\zeta{(3)}^2+\frac{75}{512}{\pi^7})$$ Using functions psi.