一个Log-Tan积分

\[\Large\int_{0}^{\pi }\theta \ln\tan\frac{\theta }{2}\mathrm{d}\theta \]

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\(\Large\mathbf{Solution:}\)
显然
\[\int_{0}^{\pi }\theta \ln\tan\frac{\theta }{2}\mathrm{d}\theta =4\int_{0}^{\pi /2}x\ln \tan x\mathrm{d}x\]
利用 \(\mathbf{Lobachevskiy}\) 函数的定义

\[\mathrm{L}\left ( x \right )=-\int_{0}^{x}\ln\cos x\mathrm{d}x~,~ ~ ~ ~ ~ -\frac{\pi }{2}\leq x\leq \frac{\pi }{2}\]

所以
\[\begin{align*} \int_{0}^{\pi /2}x\ln\tan x\mathrm{d}x &= x\left [ \mathrm{L}\left ( x \right )+\mathrm{L}\left ( \frac{\pi }{2}-x \right ) \right ]_{0}^{\pi /2}-\int_{0}^{\pi /2}\left [ \mathrm{L}\left ( x \right )+\mathrm{L}\left ( \frac{\pi }{2}-x \right ) \right ]\mathrm{d}x\\ &= \left ( \frac{\pi }{2} \right )^{2}\ln 2-2\int_{0}^{\pi /2}\mathrm{L}\left ( x \right )\mathrm{d}x \end{align*}\]
再利用
\[\mathrm{L}\left ( x \right )=x\ln 2-\frac{1}{2}\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k-1}}{k^{2}}\sin 2kx\]
就能计算得
\[\begin{align*} \int_{0}^{\pi /2}\mathrm{L}\left ( x \right )\mathrm{d}x&=\frac{1}{2}\left ( \frac{\pi }{2} \right )^{2}\ln 2-\frac{1}{2}\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k-1}}{k^{2}}\int_{0}^{\pi /2}\sin 2kx\mathrm{d}x \\ &= \frac{\pi ^{2}}{8}\ln 2-\frac{1}{2}\sum_{k=1}^{\infty }\frac{1}{\left ( 2k-1 \right )^{3}} \end{align*}\]
所以
\[\begin{align*} \int_{0}^{\pi/2}x\ln\tan x\mathrm{d}x &=\frac{\pi ^{2}}{4}\ln 2-2\left [ \frac{\pi ^{2}}{8}\ln 2-\frac{1}{2}\sum_{k=1}^{\infty }\frac{1}{\left ( 2k-1 \right )^{3}} \right ] \\ &=\sum_{k=1}^{\infty }\frac{1}{\left ( 2k-1 \right )^{3}}\\ &=\sum_{k=1}^{\infty } \frac{1}{k^{3}}-\sum_{k=1}^{\infty }\frac{1}{\left ( 2k \right )^{3}}=\frac{7}{8}\zeta \left ( 3 \right ) \end{align*}\]
亦即
\[\Large\boxed{\displaystyle \int_{0}^{\pi }\theta \ln\tan\frac{\theta }{2}\mathrm{d}\theta=\color{blue}{\frac{7}{2}\zeta \left ( 3 \right )}}\]
另外还可以得到
\[\int_{0}^{\pi }\theta \ln\tan\frac{\theta }{2}\mathrm{d}\theta=\sum_{n=1}^{\infty }\frac{1}{n^{2}}\left [ \psi \left ( n+\frac{1}{2} \right )-\psi \left ( \frac{1}{2} \right ) \right ]=\frac{7}{2}\zeta \left ( 3 \right )\]
或
\[\Large\color{purple}{\sum_{n=1}^{\infty }\frac{1}{n^{2}}\psi \left ( n+\frac{1}{2} \right )=\frac{7}{2}\zeta \left ( 3 \right )-\left ( \gamma +2\ln 2 \right )\frac{\pi ^{2}}{6}}\]

转载于:https://www.cnblogs.com/Renascence-5/p/5436780.html