Logarithmic-Trigonometric积分系列(一)

\[\Large\displaystyle \int_{0}^{\frac{\pi }{2}}x^{2}\ln\left ( \sin x \right )\ln\left ( \cos x \right )\mathrm{d}x \]


\(\Large\mathbf{Solution:}\)
Tools Needed

\[\frac{1}{k\left ( j- k \right )^{2}}=\frac{1}{j^{2}k}-\frac{1}{j^{2}\left ( k- j \right )}+\frac{1}{j\left ( k- j \right )^{2}} \]

\[\frac{1}{k\left ( j+ k \right )^{2}}=\frac{1}{j^{2}k}-\frac{1}{j^{2}\left ( k+ j \right )}-\frac{1}{j\left ( k+j \right )^{2}} \]

\[\ln\left ( \sin x \right )=-\ln 2-\sum_{k=1}^{\infty }\frac{\cos\left ( 2kx \right )}{k} \]

\[\ln\left ( \cos x \right )=-\ln 2-\sum_{k=1}^{\infty }\left ( -1 \right )^{k}\frac{\cos\left ( 2kx \right )}{k} \]

\[\cos\left ( 2jx \right )\cos\left ( 2kx \right )=\frac{1}{2}\left [ \cos\left ( 2\left ( j-k \right )x \right )+\cos\left ( 2\left ( j+k \right )x \right ) \right ] \]

\[\int_{0}^{\frac{\pi }{2}}x^{2}\cos\left ( 2kx \right )\mathrm{d}x=\begin{cases} \left ( -1 \right )^{k}\displaystyle \frac{\pi }{4k^{2}}& \text{ if } k\neq 0 \\ \displaystyle \frac{\pi ^{3}}{24}& \text{ if } k=0 \end{cases}\]

Tool Use

\[\begin{align*} &\int_{0}^{\frac{\pi }{2}}x^{2}\ln\left ( \sin x \right )\ln\left ( \cos x \right )\mathrm{d}x \\ &=\int_{0}^{\frac{\pi }{2}}x^{2}\left ( \ln 2+\sum_{k=1}^{\infty }\frac{\cos\left ( 2kx \right )}{k} \right )\left ( \ln 2+\sum_{k=1}^{\infty }\left ( -1 \right )^{k}\frac{\cos\left ( 2kx \right )}{k} \right )\mathrm{d}x \\ &=\ln^{2}2 \int_{0}^{\frac{\pi }{2}}x^{2}\mathrm{d}x+\ln 2\sum_{k=1}^{\infty }\frac{1}{k}\int_{0}^{\frac{\pi }{2}}x^{2}\cos\left ( 4kx \right )\mathrm{d}x\\ &~~~+\sum_{j=1}^{\infty }\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{2jk}\int_{0}^{\frac{\pi }{2}}x^{2}\left [ \cos\left ( 2\left ( j-k \right )x \right )+\cos\left ( 2\left ( j+k \right )x \right ) \right ]\mathrm{d}x \\ &=\frac{\pi ^{3}}{24}\ln^{2}2+\ln 2\frac{\pi }{16}\zeta \left ( 3 \right ) \\ &~~~+\frac{\pi }{8}\sum_{j=1}^{\infty }\frac{\left ( -1 \right )^{j}}{j}\sum_{k=1}^{j-1}\frac{1}{k\left ( j-k \right )^{2}}+\frac{\pi }{8}\sum_{j=1}^{\infty }\frac{\left ( -1 \right )^{j}}{j^{2}}\frac{\pi ^{2}}{6}+\frac{\pi }{8}\sum_{j=1}^{\infty }\frac{\left ( -1 \right )^{j}}{j}\sum_{k=j+1}^{\infty }\frac{1}{k\left ( j-k \right )^{2}} \\ &~~~+\frac{\pi }{8}\sum_{j=1}^{\infty }\frac{\left ( -1 \right )^{j}}{j}\sum_{k=1}^{\infty }\frac{1}{k\left ( j+k \right )^{2}} \\ &=\frac{\pi ^{3}}{24}\ln^{2}2+\ln 2\frac{\pi }{16}\zeta \left ( 3 \right ) \\ &~~~+\frac{\pi }{8}\sum_{j=1}^{\infty }\frac{\left ( -1 \right )^{j}}{j}\left ( \frac{2}{j^{2}}H_{j-1}+\frac{1}{j}H_{j-1}^{\left ( 2 \right )} \right )-\frac{\pi ^{5}}{576}+\frac{\pi }{8}\sum_{j=1}^{\infty }\frac{\left ( -1 \right )^{j}}{j}\left ( -\frac{1}{j^{2}}H_{j}+\frac{1}{j}\frac{\pi ^{2}}{6} \right )\\ &~~~+\frac{\pi }{8}\sum_{j=1}^{\infty }\frac{\left ( -1 \right )^{j}}{j}\left ( \frac{1}{j^{2}}H_{j}-\frac{1}{j}\frac{\pi ^{2}}{6}+\frac{1}{j}H_{j}^{\left ( 2 \right )} \right ) \\ &=\frac{\pi ^{3}}{24}\ln^{2}2+\ln 2\frac{\pi }{16}\zeta \left ( 3 \right ) \\ &~~~+\frac{\pi }{8}\sum_{j=1}^{\infty }\frac{\left ( -1 \right )^{j}}{j}\left ( \frac{2}{j^{2}}H_{j}+\frac{2}{j}H_{j}^{\left ( 2 \right )}-\frac{3}{j^{3}} \right )-\frac{\pi ^{5}}{576} \\ &=\frac{\pi ^{3}}{24}\ln^{2}2+\ln 2\frac{\pi }{16}\zeta \left ( 3 \right )+\frac{11\pi ^{5}}{5760}+\frac{\pi }{4}\sum \left ( -1 \right )^{j}\left ( \frac{1}{j^{3}}H_{j}+\frac{1}{j^{2}}H_{j}^{\left ( 2 \right )} \right ) \\ &=\frac{\pi ^{3}}{24}\ln^{2}2+\ln 2\frac{\pi }{16}\zeta \left ( 3 \right )-\frac{\pi ^{5}}{960}-\frac{\pi }{16}\sum_{j=1}^{\infty }\frac{H_{2j}}{j^{3}} \end{align*}\]

Using the known result

\[\sum_{n=1}^{\infty }\frac{H_{2n}}{n^{3}}=-\frac{\pi ^{4}}{15}-\frac{1}{3}\pi ^{2}\ln^{2}2+\frac{\ln^{4}2}{3}+8\mathrm{Li}_{4}\left ( \frac{1}{2} \right )+7\ln 2\zeta \left ( 3 \right ) \]

So here is the final result:

\[\Large\boxed{\displaystyle \begin{align*} \int_{0}^{\frac{\pi }{2}}x^{2}\ln\left ( \sin x \right )\ln\left ( \cos x \right )\mathrm{d}x&=\color{blue}{\frac{\pi ^{3}}{16}\ln^{2}2+\frac{\pi ^{5}}{320}-\frac{3}{8}\ln 2\zeta \left ( 3 \right )}\\ &~~~\color{blue}{-\frac{\pi }{48}\ln^{4}2-\frac{1}{2}\mathrm{Li}_{4}\left ( \frac{1}{2} \right )} \end{align*}}\]

posted @ 2016-05-13 10:48  Renascence_5  阅读(329)  评论(0编辑  收藏