Logarithmic-Trigonometric积分系列（二）-优快云博客

\[\Large\displaystyle \int_0^{\pi/2}\ln^2(\sin x)\ln(\cos x)\tan x \,{\rm d}x\]

\(\Large\mathbf{Solution:}\)
Let \(J\) donates the integral and it is easy to see that
\[\begin{align*} J&=\int_0^{\pi/4}\ln^2(\sin x)\ln(\cos x)\tan x \,{\rm d}x+ \int_{\pi/4}^{\pi/2}\ln^2(\sin x)\ln(\cos x)\tan x \,{\rm d}x\cr &=\int_0^{\pi/4}\ln^2(\sin x)\ln(\cos x)\tan x \,{\rm d}x+ \int_{0}^{\pi/4}\ln^2(\cos x)\ln(\sin x)\cot x \,{\rm d}x\cr \end{align*}\]
Now, to calculate \(J\) we make the substitution \(t\leftarrow\sin^2x\):
\[J=\frac{1}{16}\int_0^1\frac{\ln(1-u)}{1-u}\ln^2(u)\,{\rm d}u\]
But
\[\frac{\ln(1-u)}{1-u}=-\left(\sum_{n=0}^\infty u^n\right)\left(\sum_{n=1}^\infty \frac{u^n}{n}\right) =-\sum_{n=1}^\infty H_nu^n\]
where \(H_n=\displaystyle\sum_{k=1}^n \frac{1}{k}\).Hence
\[J=-\frac{1}{16}\sum_{n=1}^\infty H_n\int_0^1u^n\ln^2(u){\rm d}u =-\frac{1}{8}\sum_{n=1}^\infty\frac{ H_n}{(n+1)^3}\]
The sum \(\displaystyle\sum_{n=1}^\infty\frac{H_n}{n^3}\) is known, it can be evaluated as follows, first we have
\[H_n=\sum_{k=1}^\infty\left(\frac{1}{k}-\frac{1}{k+n}\right)= \sum_{k=1}^\infty \frac{n}{k(k+n)}\]
Thus
\[\sum_{n=1}^\infty\frac{H_n}{n^3}=\sum_{k,n\geq1}\frac{1}{n^2k(n+k)} =\sum_{k,n\geq1}\frac{1}{k^2n(n+k)}\]
Taking the half sum we find
\[\sum_{n=1}^\infty\frac{H_n}{n^3}=\frac{1}{2}\sum_{k,n\geq1}\frac{1}{kn(k+n)}\left(\frac{1}{k}+\frac{1}{n}\right)= \frac{1}{2}\sum_{k,n\geq1}\frac{1}{k^2n^2}=\frac{1}{2}\zeta^2(2)\]
then we obtain
\[\Large\boxed{\displaystyle \begin{align*} \int_0^{\pi/2}\ln^2(\sin x)\ln(\cos x)\tan x \,{\rm d}x&=\frac{1}{8}\zeta(4)-\frac{1}{16}\zeta^2(2)\\ &=\color{blue}{-\frac{\pi^4}{2880}} \end{align*}}\]