# Logarithmic-Trigonometric积分系列（二）

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

$\Large\mathbf{Solution:}$
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*}}

posted @ 2016-05-13 10:58  Renascence_5  阅读(298)  评论(0编辑  收藏