复杂的对数积分(八)

\[\Large\displaystyle \int_0^\infty\frac{\ln\left(1+x+\sqrt{x^2+2\,x}\right)\,\ln\left(1+\sqrt{x^2+2\,x+2}\right)}{x^2+2x+1}\mathrm dx \]


\(\Large\mathbf{Solution:}\)
This integral can be solved by brutal force.

\[I = \int_{1}^{\infty} \frac{\ln(x + \sqrt{x^{2} - 1}) \ln(1 + \sqrt{x^{2} + 1})}{x^{2}} \, \mathrm dx \]

By the knowledge in trigonometry, we obtain

\[\begin{align*} I &= \int_{1}^{\infty} \frac{\operatorname{arcosh} x \cdot (\operatorname{arsinh}(1/x) + \ln x)}{x^{2}} \, \mathrm dx \\ &= \int_{0}^{1} \operatorname{arcosh} (1/x) \cdot (\operatorname{arsinh} x - \ln x) \, \mathrm dx \end{align*}\]

Note that

\[\int (\operatorname{arsinh} x - \ln x) \, \mathrm dx = x\operatorname{arsinh} x - x \ln x + x + 1 - \sqrt{x^{2} + 1} \]

Here, the constant of integration is chosen so that the integrand becomes \(O(x\ln x)\) as \(x \searrow 0\). Since \(\operatorname{arcosh}(1/x) = O(\ln x)\) as \(x \searrow 0\), we can perform integration by parts to obtain

\[I = \int_{0}^{1} \left( \operatorname{arsinh} x - \ln x + 1 + \frac{1 - \sqrt{x^{2} + 1}}{x} \right) \, \frac{\mathrm dx}{\sqrt{1-x^{2}}} \]

Plug \(x=\sin\theta\). Then

\[I = \int_{0}^{\frac{\pi}{2}} \left( \operatorname{arsinh} (\sin\theta) - \ln \sin\theta + 1 + \frac{1 - \sqrt{1 + \sin^{2}\theta}}{\sin\theta} \right) \, \mathrm d\theta \]

We divide the integral into 4 parts and consider them separately.
Part 1. By the Taylor expansion of \(\operatorname{arsinh} x\),

\[\begin{align*} \int_{0}^{\frac{\pi}{2}} \operatorname{arsinh} (\sin\theta) \, \mathrm d\theta &= \sum_{n=0}^{\infty} \binom{-1/2}{n} \frac{1}{2n+1} \int_{0}^{\frac{\pi}{2}} \sin^{2n+1}\theta \, \mathrm d\theta \\ &= \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{2n+1} \frac{\Gamma\left ( \dfrac{1}{2} \right )}{\Gamma(n+1)\Gamma\left ( \dfrac{1}{2} \right )} \frac{\Gamma(n+1)\Gamma\left ( \dfrac{1}{2} \right )}{\Gamma\left ( \dfrac{3}{2}+n \right )} \\ &= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}} \\&= \textbf{G} \end{align*}\]

Part 2. We know so many ways to prove that

\[-\int_{0}^{\frac{\pi}{2}} \ln \sin\theta \, \mathrm d\theta = \frac{\pi}{2}\ln 2 \]

Part 3. Do you need an explanation for this?

\[\int_{0}^{\frac{\pi}{2}} \mathrm d\theta = \frac{\pi}{2} \]

Part 4. Again, by the Taylor expansion,

\[\begin{align*} \int_{0}^{\frac{\pi}{2}} \frac{1 - \sqrt{1 + \sin^{2}\theta}}{\sin\theta} \, \mathrm d\theta &= - \sum_{n=1}^{\infty} \binom{1/2}{n} \int_{0}^{\frac{\pi}{2}} \sin^{2n-1}\theta \, \mathrm d\theta \\ &= - \frac{1}{2} \sum_{n=1}^{\infty} \frac{\Gamma\left ( \displaystyle\frac{3}{2} \right )}{\Gamma(n+1)\Gamma\left ( \displaystyle\frac{3}{2}-n \right )} \frac{\Gamma(n)\Gamma\left ( \displaystyle\frac{1}{2} \right )}{\Gamma\left ( \displaystyle\frac{1}{2}+n \right )} \\ &= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n)(2n-1)} \\&= \frac{\ln 2}{2} - \frac{\pi}{4} \end{align*}\]

Putting all these together, we obtain

\[\boxed{\displaystyle \int_0^\infty\frac{\ln\left(1+x+\sqrt{x^2+2\,x}\right)\,\ln\left(1+\sqrt{x^2+2\,x+2}\right)}{x^2+2x+1}\mathrm dx=\color{Blue} {\textbf{G}+\frac\pi4+\frac{\pi+1}2\ln2}} \]

posted @ 2016-05-14 21:52  Renascence_5  阅读(297)  评论(0编辑  收藏