Evaluating $\int_{0}^{\pi/4} \log(\sin(x)) \log(\cos(x)) \log(\cos(2x)) \,dx$

Question

What tools would you recommend me for evaluating this integral?

$$\int_{0}^{\pi/4} \log(\sin(x)) \log(\cos(x)) \log(\cos(2x)) \,dx$$

My first thought was to use the beta function, but it's hard to get such a form because
of $\cos(2x)$. What other options do I have?

Answer 1

A decade soon, and still no solution! (let's put an end to this story - very elegantly)

A solution by Cornel Ioan Valean (in large steps)

To begin with, I'll consider the trivial integral result, $$ \int_0^{\pi/4} \sin^2(2 n x) \log (\cos (2 x)) \textrm{d}x=-\log(2)\frac{\pi}{8}-\frac{\pi}{16} (-1)^{n-1}\frac{1}{n},$$ which is straightforward to extract by using the identity $1-\cos(x)=2\sin^2(x/2)$, integrating the trivial resulting part, then integrating by parts the remaining integral and combining it with the use of a simple recurrence relation (also given in the sequel, page $201$).

The key step is represented by the Fourier-like series presented and derived in the book (Almost) Impossible Integrals, Sums, and Series (2019), pages $248$, $$\small \sum_{n=1}^{\infty} \left(2H_{2n}-2H_n+\frac{1}{2n}-2\log(2)\right)\frac{\sin^2(2nx)}{n}=\log(\sin(x))\log(\cos(x)), \ 0< x<\frac{\pi}{2}.$$

Returning to the main integral, we have that $$\int_0^{\pi/4} \log(\sin(x))\log(\cos(x))\log(\cos(2x)) \textrm{d}x$$ $$=\sum_{n=1}^{\infty} \left(2H_{2n}-2H_n+\frac{1}{2n}-2\log(2)\right)\frac{1}{n} \int_0^{\pi/4} \sin^2(2nx) \log(\cos(2x)) \textrm{d}x$$ $$=-\sum_{n=1}^{\infty} \left(2H_{2n}-2H_n+\frac{1}{2n}-2\log(2)\right)\frac{1}{n} \left(\log(2)\frac{\pi}{8}+\frac{\pi}{16} (-1)^{n-1}\frac{1}{n}\right)$$ $$\small =-\frac{3}{128}\pi \zeta(3)-\log(2)\frac{\pi}{4}\sum_{n=1}^{\infty} \frac{1}{n}\left(H_{2n}-H_n-\log(2)\right) +\frac{\pi}{8} \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n}{n^2}-\frac{\pi}{8} \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n^2}$$ $$=\frac{\pi^3}{48}\log(2)-\frac{\pi}{4}\log^3(2)+\frac{15}{64}\pi \zeta(3)-\frac{\pi^2}{8}G,$$ where all the series are known. For example, we have that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}\left(H_{2n}-H_n-\log(2)\right)=\log^2(2)-\frac{\pi^2}{12}$, which is given and calculated both in (Almost) Impossible Integrals, Sums, and Series (2019), page $250$, and its sequel More (Almost) Impossible Integrals, Sums, and Series (2023), page $767$, then $\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n}{n^2}=\frac{5}{8}\zeta(3)$ can be found both in the first mention book, page $310$, but also in the sequel, where you can find solutions by elementary series manipulations, which means you won't need to touch integrals (check pages $414$ and $421$ and the related solution sections). At the same time, you can also exploit generating functions to get the result, again in the sequel (page $399$). Further, using the generating function stated at eq. $(4.36)$ (as previously suggested), with the third closed-form, we get that $\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n^2}=-4 \Re \sum_{n=1}^{\infty} i^n \frac{H_n}{n^2}=\pi G-\frac{23}{16}\zeta(3)$, where the resulting dilogarithmic and trilogarithmic values with the needed complex argument can be found in the sequel, the section $6.53$, pages $770$-$779$ and also here. Another thing to emphasize at this point is that we can totally avoid the use of complex numbers for the series $\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n^2}$ due to the fact that in (Almost) Impossible Integrals, Sums, and Series (2019), pages $123$-$124$, the related integral, that is, $\displaystyle \int_0^1 \frac{x \operatorname{Li}_2(x)}{1+x^2}\textrm{d}x$ is calculated by real methods (the calculations can be optimized even more by exploiting results from the sequel).

What's left? (since we are done with the main integral)

Some temptations for adventurous calculations may immediately cross one's mind if we recall the splendid Fourier series,

$$\log(\cos(x))\log^2(\sin(x))$$ $$=\frac{1}{4}\zeta(3)-\log^3(2)+\sum _{n=1}^{\infty } \biggr(\int_0^1 t^{n-1} \biggr(4 \operatorname{arctanh}^2(t)-\log ^2\left(\frac{1-t}{2 \sqrt{t}}\right)$$ $$-\left(1-(-1)^{n-1}\right) \log ^2\left(\frac{1+t}{2 \sqrt{t}}\right)\biggr)\textrm{d}t \biggr)\cos (2 n x), \ 0<x<\frac{\pi}{2};$$ or $$\log(\sin(x))\log^2(\cos(x))$$ $$=\frac{1}{4}\zeta(3)-\log^3(2)-\sum _{n=1}^{\infty }(-1)^{n-1} \biggr(\int_0^1 t^{n-1} \biggr(4 \operatorname{arctanh}^2(t)-\log ^2\left(\frac{1-t}{2 \sqrt{t}}\right)$$ $$-\left(1-(-1)^{n-1}\right) \log ^2\left(\frac{1+t}{2 \sqrt{t}}\right)\biggr)\textrm{d}t \biggr)\cos (2 n x), \ 0<x<\frac{\pi}{2},$$ which are given in More (Almost) Impossible Integrals, Sums, and Series (2023), page $449$.

End of story (by the way, soon more advanced versions of this integral will appear in a paper)

Answer 2

The function $g(x)=\left(1-\left(\frac{4x}{\pi }\right)^4\right)^{\frac{1}{4}}-1$ is a good approximation of the integrand. And its integral between $[0,\frac{\pi}{4}]$ is equal to $\frac{1}{8}(\sqrt{\pi}\Gamma(\frac{1}{4})\Gamma(\frac{5}{4})-2\pi)$ which approximatly equals to $-0.0573047$.

Answer 3

Mathematica cannot find an expression of this integral in terms of elementary functions. However, it can be integrated numerically, like this

 NIntegrate[Log[Sin[x]] Log[Cos[x]] Log[Cos[2 x]], {x, 0, Pi/4}]

to yield the result

 -0.05874864

Edit: I'm told an analytic expression might still exist, even though it cannot be found with Mathematica. Regardless, the numeric value is the one above.