# Delightful integral: $\int _0^{\frac{\pi }{2}}x\ln \left(\sin \left(x\right)\right)\ln ^2\left(\cos \left(x\right)\right)\:dx.$

A friend of mine proposed me this integral which I find to be very interesting.

I managed to find with the help of software that: $$\int _0^{\frac{\pi }{2}}x\ln \left(\sin \left(x\right)\right)\ln ^2\left(\cos \left(x\right)\right)\:dx=\frac{155}{128}\zeta \left(5\right)+\frac{13}{32}\zeta \left(2\right)\zeta \left(3\right)-\operatorname{Li}_5\left(\frac{1}{2}\right)$$ $$-\frac{49}{32}\ln \left(2\right)\zeta \left(4\right)-\frac{5}{6}\ln ^3\left(2\right)\zeta \left(2\right)+\frac{1}{120}\ln ^5\left(2\right).$$ Where $$\zeta \left(z\right)$$ denotes the Riemann zeta function and $$\operatorname{Li}_n\left(z\right)$$ denotes the Polylogarithm function.

If $$x=\tan\left(t\right)$$ is used it yields: $$\frac{1}{4}\int _0^{\infty }\frac{\arctan \left(x\right)\ln \left(x\right)\ln ^2\left(1+x^2\right)}{1+x^2}\:dx-\frac{1}{8}\int _0^{\infty }\frac{\arctan \left(x\right)\ln ^3\left(1+x^2\right)}{1+x^2}\:dx.$$ I know of ways to evaluate the latter integral but the former is very difficult and the techniques that work for the $$2$$nd do not work for the $$1$$st.

Integrating by parts gives: $$-\frac{1}{2}\int _0^{\frac{\pi }{2}}x^2\cot \left(x\right)\ln ^2\left(\cos \left(x\right)\right)\:dx-\int _0^{\frac{\pi }{2}}x^2\tan \left(x\right)\ln \left(\sin \left(x\right)\right)\ln \left(\cos \left(x\right)\right)\:dx.$$ And I find myself in the same situation where I can evaluate the $$1$$st integral but the techniques that work for it aren't as effective for the $$2$$nd, are there any better approaches for the main integral?

Please do not post results without proving them since that is not what I'm looking for, thanks.

• I know a way to evaluate the original integral, also to explain why only $\text{Li}_n(1/2)$ appear on RHS, but not Catalan constant or $\beta(n)$ or $\text{Li}_n(\frac{1+i}{2})$ for example. The method is not simple, but can be immediately generalized to $$\int_0^{\pi/2} x^a \log^b(\sin x) \log^c(\cos x) dx$$ Let me think if my method can be articulated using techniques known to the OP, who I believe is a reader of Prof. Cornel Vălean's books. Jun 28, 2021 at 8:36
• @cleo ${}{}{}{}{}$ Jun 30, 2021 at 0:28
• @pisco How did you get so good at integration? I've noticed that you almost always have solutions to really tough integrals. Jul 1, 2021 at 20:57
• @A-LevelStudent Thank you, that's certainly an overrated and exaggerated compliment. Just like any skill, practice is paramount. I have been interested in this area for quite a long time, initially I could only do simple problems, but one will be more powerful as time progress. For more concrete advise (regarding integral & series in general), you can proceed via following stages (1) differentiating under integral / exchange of integral and series; (2) basic complex analysis (in textbooks), especially contour integration; (3) basic special functions (e.g. gamma, zeta, polylogarithm); Jul 2, 2021 at 2:05
• ... (4) more on contour integration (beyond textbook examples), you can find daunting integrals that can be solved (and possibly only solved) via this method on MSE; (5) more special functions (e.g. Bessel, hypergeometric, elliptic and theta); (6) Mellin/Laplace/Fourier transform and their inversion; Of course this is a very subjective suggestion, and there are far more advanced techniques and topics which I am still learning. Nonetheless steps (1)-(3) will do good to beginners. Jul 2, 2021 at 2:10

$$I=\int _0^{\frac{\pi }{2}}x\ln \left(\sin \left(x\right)\right)\ln ^2\left(\cos \left(x\right)\right)\:dx=\frac{155}{128}\zeta \left(5\right)+\frac{13}{32}\zeta \left(2\right)\zeta \left(3\right)-\operatorname{Li}_5\left(\frac{1}{2}\right)$$ $$-\frac{49}{32}\ln \left(2\right)\zeta \left(4\right)-\frac{5}{6}\ln ^3\left(2\right)\zeta \left(2\right)+\frac{1}{120}\ln ^5\left(2\right),$$ $$J=\int _0^{\frac{\pi }{2}}x\ln ^2\left(\sin \left(x\right)\right)\ln \left(\cos \left(x\right)\right)\:dx=-\frac{155}{128}\zeta \left(5\right)-\frac{1}{32}\zeta \left(2\right)\zeta \left(3\right)+\operatorname{Li}_5\left(\frac{1}{2}\right)$$ $$+\frac{49}{32}\ln \left(2\right)\zeta \left(4\right)-\frac{2}{3}\ln ^3\left(2\right)\zeta \left(2\right)-\frac{1}{120}\ln ^5\left(2\right).$$
Let's consider calculating $$I-J$$ first. $$I-J=\int _0^{\frac{\pi }{2}}x\ln \left(\sin \left(x\right)\right)\ln ^2\left(\cos \left(x\right)\right)\:dx-\int _0^{\frac{\pi }{2}}x\ln ^2\left(\sin \left(x\right)\right)\ln \left(\cos \left(x\right)\right)\:dx$$ $$=\frac{1}{3}\underbrace{\int _0^{\infty }\frac{\ln ^3\left(x\right)\arctan \left(x\right)}{1+x^2}\:dx}_{\mathcal{A}}-\frac{2}{3}\underbrace{\int _0^{\frac{\pi }{2}}x\ln ^3\left(\sin \left(x\right)\right)\:dx}_{\mathcal{B}}+\frac{\pi }{6}\underbrace{\int _0^{\frac{\pi }{2}}\ln ^3\left(\sin \left(x\right)\right)\:dx}_{\mathcal{C}}.$$ Here $$\mathcal{A}$$ and $$\mathcal{C}$$ are very straightforward, the former can be easily calculated through differentiation under the integral sign considering the following parameter: $$\mathcal{A}\left(a\right)=\int _0^{\infty }\frac{\ln ^3\left(x\right)\arctan \left(ax\right)}{1+x^2}\:dx.$$ After the usual procedure it reduces to trivial/well-known integrals, in the other hand $$\mathcal{C}$$ may be calculated employing the Beta function, so: $$\mathcal{A}=\int _0^{\infty }\frac{\ln ^3\left(x\right)\arctan \left(x\right)}{1+x^2}\:dx=\frac{93}{16}\zeta \left(5\right)+\frac{21}{16}\zeta \left(2\right)\zeta \left(3\right),$$ $$\mathcal{C}=\int _0^{\frac{\pi }{2}}\ln ^3\left(\sin \left(x\right)\right)\:dx=-\frac{3\pi }{4}\zeta \left(3\right)-\frac{\pi ^3}{8}\ln \left(2\right)-\frac{\pi }{2}\ln ^3\left(2\right).$$ This only leaves us with the toughest one which is: $$\mathcal{B}=\int _0^{\frac{\pi }{2}}x\ln ^3\left(\sin \left(x\right)\right)\:dx=-\frac{\pi }{16}\underbrace{\int _0^{\infty }\frac{\ln ^3\left(1+x^2\right)}{1+x^2}\:dx}_{\operatorname{Beta function}}-\frac{1}{8}\int _0^{\infty }\frac{\arctan \left(\frac{1}{x}\right)\ln ^3\left(\frac{x^2}{1+x^2}\right)}{1+x^2}\:dx$$ $$=-\frac{9}{4}\zeta \left(2\right)\zeta \left(3\right)-\frac{45}{8}\ln \left(2\right)\zeta \left(4\right)-\frac{3}{2}\ln ^3\left(2\right)\zeta \left(2\right)-\frac{1}{4}\operatorname{\mathfrak{I}} \left\{\int _0^{\infty }\frac{\ln ^4\left(\frac{x}{x-i}\right)}{1+x^2}\:dx\right\}$$ $$+\frac{1}{2}\int _0^{\infty }\frac{\arctan ^3\left(x\right)\ln \left(1+x^2\right)}{1+x^2}\:dx$$ $$=-\frac{9}{4}\zeta \left(2\right)\zeta \left(3\right)-\frac{45}{8}\ln \left(2\right)\zeta \left(4\right)-\frac{3}{2}\ln ^3\left(2\right)\zeta \left(2\right)$$ $$-\frac{1}{4}\operatorname{\mathfrak{I}} \left\{-12i\operatorname{Li}_5\left(2\right)+\pi \ln ^4\left(2\right)\right\}-\int _0^{\frac{\pi }{2}}x^3\ln \left(\cos \left(x\right)\right)\:dx.$$ Where the fourier series of $$\ln \left(\cos \left(x\right)\right)$$ may be employed for that last integral for a fast calculation. In the end we obtain that: $$\mathcal{B}=\int _0^{\frac{\pi }{2}}x\ln ^3\left(\sin \left(x\right)\right)\:dx=-\frac{93}{128}\zeta \left(5\right)-\frac{9}{8}\zeta \left(2\right)\zeta \left(3\right)+3\operatorname{Li}_5\left(\frac{1}{2}\right)$$ $$+\frac{57}{32}\ln \left(2\right)\zeta \left(4\right)-\frac{1}{2}\ln ^3\left(2\right)\zeta \left(2\right)-\frac{1}{40}\ln ^5\left(2\right).$$ Collecting all the previous results leads us to the closed-form of $$I-J$$ which is: $$I-J=\int _0^{\frac{\pi }{2}}x\ln \left(\sin \left(x\right)\right)\ln ^2\left(\cos \left(x\right)\right)\:dx-\int _0^{\frac{\pi }{2}}x\ln ^2\left(\sin \left(x\right)\right)\ln \left(\cos \left(x\right)\right)\:dx$$ $$=\frac{155}{64}\zeta \left(5\right)+\frac{7}{16}\zeta \left(2\right)\zeta \left(3\right)-2\operatorname{Li}_5\left(\frac{1}{2}\right)-\frac{49}{16}\ln \left(2\right)\zeta \left(4\right)-\frac{1}{6}\ln ^3\left(2\right)\zeta \left(2\right)+\frac{1}{60}\ln ^5\left(2\right).$$ Note that if we apply the substitution $$x=\frac{\pi }{2}-t$$ to either $$I$$ or $$J$$ we can easily find each of them separately.