# Closed form $\int_{0}^{\pi/2}{\ln\cos x\over \tan^sx}\cdot\ln\left({\ln^2(\cos x)\over \pi^2+\ln^2(\sin x)}\right)\mathrm dx$

Consider $(1)$, once again the motivation of this question is from Integral contest

$$\int _0^{\frac{\pi }{2}}\frac{\ln \left(\cos x\right)}{\tan^s(x)}\ln \left(\frac{\ln ^2\left(\cos x\right)}{\pi ^2+\ln ^2\left(\sin x\right)}\right)\mathrm dx=F(s)\tag1$$ Where $s\ge1$

I noticed that $$F(1)=\zeta(2)\tag2$$

$$F(2)=3\zeta(3)\tag3$$

How may one find the closed form of $F(s)?$

• $\ln^2 \cos^2 x=(\ln((\cos x)^2))^2$ ? – FDP Apr 15 '17 at 17:46
• @FDP If this is what the OP meant, in Desmos I numerically get $1.10316... \neq \zeta(2) = \frac{\pi^2}{6} = 1.6449...$ Here is a link confirming this with WA – Brevan Ellefsen Apr 15 '17 at 18:25
• @FDP after playing around with it for a while, I think the OP meant $$\int _0^{\frac{\pi }{2}}\frac{\ln \left(\cos x\right)}{\tan \left(x\right)}\ln \left(\frac{\ln ^2\left(\cos x\right)}{\pi ^2+\ln ^2\left(\sin x\right)}\right)dx$$ – Brevan Ellefsen Apr 15 '17 at 18:40
• I am sorry for the mistake posting thank you so much editing it – gymbvghjkgkjkhgfkl Apr 15 '17 at 21:02
• @Latte' As $x \to 0^+$, the integrand is such that $$\frac{\ln \left(\cos x\right)}{\tan^s \left(x\right)}\ln \left(\frac{\ln ^2\left(\cos x\right)}{\pi ^2+\ln ^2\left(\sin x\right)}\right) \sim- \frac{2\ln x}{x^{s-2}}$$ yielding that the given integral is divergent for $s-2\ge1$ that is for $s\ge3$. Thus $F(s)$ does not exist for $s\ge3$. – Olivier Oloa Apr 15 '17 at 22:55