复杂的对数积分（六）

\[\Large\displaystyle \int_0^1 \dfrac{\operatorname{Li}_2\left(\dfrac{x}{4}\right)}{4-x}\ln\left(\dfrac{1+\sqrt{1-x}}{1-\sqrt{1-x}}\right)\mathrm{d}x=\dfrac{\pi^4}{1944}\]

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\(\Large\mathbf{Proof:}\)
\[\begin{align*} {\int_0^1 \frac{\text{Li}_2\left(\dfrac{x}{4}\right)}{4-x}\ln\left(\frac{1+\sqrt{1-x}}{1-\sqrt{1-x}} \right) \mathrm{d}x} &= \frac{1}{4}\int_0^1\left(\sum_{n=1}^\infty\frac{H_{n}^{(2)}x^n}{2^{2n}} \right)\ln\left(\frac{1+\sqrt{1-x}}{1-\sqrt{1-x}} \right) \mathrm{d}x \\ &= \sum_{n=1}^\infty\frac{H_n^{(2)}}{2^{2(n+1)}}\int_0^1 x^n \ln\left(\frac{1+\sqrt{1-x}}{1-\sqrt{1-x}} \right) \mathrm{d}x\\&=\sum_{n=1}^\infty\frac{H_n^{(2)}}{(n+1)2^{2(n+1)}}\int_0^1\frac{x^n}{\sqrt{1-x}}\mathrm{d}x \\ &= \frac{1}{2}\sum_{n=1}^\infty\frac{H_n^{(2)}}{(n+1)(2n+1)\dbinom{2n}{n}}\\&=\sum_{n=1}^\infty\frac{H_n^{(2)}}{(n+1)^2\dbinom{2(n+1)}{n+1}}\\&=\frac{2}{3}\left(\sin^{-1}\frac{1}{2} \right)^4 \\ &=\Large\boxed{\color{blue}{\dfrac{\pi^4}{1994}}} \end{align*}\]

转载于:https://www.cnblogs.com/Renascence-5/p/5475265.html