# 一个双曲函数的积分

$\Large\displaystyle \int^{\infty}_{0}\frac{\tanh\left(\, x\,\right)} {x\left[\, 1 - 2\cosh\left(\, 2x\,\right)\,\right]^{2}}\,{\rm d}x$

$\Large\mathbf{Solution:}$
A possible way I see of doing this is to apply the substitution $x\mapsto-\ln{x}$, which yields

$-\int^{1}_{0}\frac{x^{3}\left(\, 1 - x^{2}\,\right)} {\left(\, 1 + x^{2}\,\right)\left(\, 1 - x^{2} + x^{4}\,\right)^{2}}\, \frac{{\rm d}x}{\ln\left(\, x\,\right)}$

So

\begin{align*} \mathcal{I} &=-\int_{0}^{1}\frac{x^3(1-x^2)}{(1+x^2)(1-x^2+x^4)^2}\frac{\mathrm{d}x}{\ln{x}}\\&=-\int_{0}^{1}\frac{x^3(1-x^2)}{(1+x^6)(1-x^2+x^4)}\frac{\mathrm{d}x}{\ln{x}}\\ &=-\int_{0}^{1}\frac{x^3(1-x^4)}{(1+x^6)^2}\frac{\mathrm{d}x}{\ln{x}}\\&=-\int_{0}^{1}\frac{z(1-z^{4/3})}{(1+z^2)^2}\frac{\mathrm{d}z}{3z^{2/3}\ln{\left(z^{1/3}\right)}}\\&=\int_{0}^{1}\,\frac{1}{(1+z^2)^2}\frac{z^{5/3}-z^{1/3}}{\ln{z}}\mathrm{d}z\\ &=\int_{0}^{1}\,\frac{\mathrm{d}z}{(1+z^2)^2}\int_{1/3}^{5/3}\,z^{\mu}\mathrm{d}\mu\\&=\int_{1/3}^{5/3}\mathrm{d}\mu\int_{0}^{1}\,\frac{z^{\mu}}{(1+z^2)^2}\mathrm{d}z\\&=\int_{1/3}^{5/3}\left[-\frac14+\frac{\mu-1}{4}\beta{\left(\frac{\mu-1}{2}\right)}\right]\mathrm{d}\mu\\ &=-\frac13+\int_{1/3}^{5/3}\left[\frac{\mu-1}{4}\beta{\left(\frac{\mu-1}{2}\right)}\right]\mathrm{d}\mu\\&=-\frac13+\int_{-1/3}^{1/3}\,t\beta{\left(t\right)}\mathrm{d}t\\ &=-\frac13+\int_{-1/3}^{1/3}\,\frac{t}{2}\left[\psi{\left(\frac{t+1}{2}\right)}-\psi{\left(\frac{t}{2}\right)}\right]\mathrm{d}t\\&=-\frac13+\int_{-1/3}^{1/3}\,\frac{t}{2}\psi{\left(\frac{t+1}{2}\right)}\mathrm{d}t-\int_{-1/3}^{1/3}\,\frac{t}{2}\psi{\left(\frac{t}{2}\right)}\mathrm{d}t\\ &=-\frac13+\int_{1/3}^{2/3}\,(2u-1)\psi{\left(u\right)}\mathrm{d}u-2\int_{-1/6}^{1/6}\,u\psi{\left(u\right)}\mathrm{d}u\\ &=-\frac13-\int_{1/3}^{2/3}\,\psi{\left(u\right)}\mathrm{d}u+2\int_{1/3}^{2/3}\,u\psi{\left(u\right)}\mathrm{d}u-2\int_{-1/6}^{1/6}\,u\psi{\left(u\right)}\mathrm{d}u\\ &=-\frac13+\ln{\left(\frac{\Gamma{\left(\dfrac13\right)}}{\Gamma{\left(\dfrac23\right)}}\right)}+2\int_{1/3}^{2/3}\,u\psi{\left(u\right)}\mathrm{d}u-2\int_{-1/6}^{1/6}\,u\psi{\left(u\right)}\mathrm{d}u\\ &=-\frac13+\ln{\left(\frac{\Gamma{\left(\dfrac13\right)}}{\Gamma{\left(\dfrac23\right)}}\right)}+2\int_{1/3}^{2/3}\,u\psi{\left(u\right)}\mathrm{d}u-2\int_{5/6}^{7/6}\,(1-v)\psi{\left(1-v\right)}\mathrm{d}v\\ &=-\frac13+\ln{\left(\frac{\Gamma{\left(\dfrac13\right)}}{\Gamma{\left(\dfrac23\right)}}\right)}+2\left[u\ln{\Gamma\left(u\right)}-\psi^{(-2)}{\left(u\right)}\right]_{1/3}^{2/3}\\ &~~~~~ +2\left[(1-v)\ln{\Gamma\left(1-v\right)}-\psi^{(-2)}{\left(1-v\right)}\right]_{5/6}^{7/6}\\ &=\ln{\left(\frac{\Gamma{\left(\dfrac13\right)}}{\Gamma{\left(\dfrac23\right)}}\right)}-\frac{5\pi}{9\sqrt{3}}-\frac{\ln{\left(2\pi\right)}}{3}+\frac23\ln{\left(\frac{\Gamma{\left(\dfrac23\right)}^2}{\Gamma{\left(\dfrac13\right)}}\right)}+\frac{5\psi^{(1)}{\left(\dfrac13\right)}}{6\sqrt{3}\,\pi}\\ &=\Large\boxed{\displaystyle\color{blue}{-\frac{5\pi}{9\sqrt{3}}-\frac{\ln{3}}{6}+\frac{5\psi^{(1)}{\left(\dfrac13\right)}}{6\sqrt{3}\,\pi}}} \end{align*}

posted @ 2016-05-15 21:06  Renascence_5  阅读(1163)  评论(0编辑  收藏