这里感谢风碎便士和西哥的提示。

1.计算
\[ \int_{0}^{\infty}{\frac{(1-e^{-6x})e^{-x}}{x(1+e^{-2x}+e^{-4x}+e^{-6x}+e^{-8x})}dx} \]
解
\begin{align*}
 &\int_{0}^{\infty}{\frac{(1-e^{-6x})e^{-x}}{x(1+e^{-2x}+e^{-4x}+e^{-6x}+e^{-8x})}dx}\\
 &= \int_{0}^{\infty}{\frac{(1-e^{-6x})e^{-x}(1-e^{-2x})}{x(1-e^{-10x})}dx}\\
 &=\int_{0}^{\infty}{\sum_{n=0}^{\infty}{ \frac{(1-e^{-6x})e^{-(10n+1)x}(1-e^{-2x})}{x}}dx}\\
 &=\int_{0}^{\infty}{\sum_{n=0}^{\infty}{\frac{(e^{-(10n+1)x}-e^{-(10n+7)x})-(e^{-(10n+3)x}-e^{-(10n+9)x})}{x}}dx}\\
 &=\sum_{n=0}^{\infty}{\ln{\frac{(10n+7)(10n+3)}{(10n+1)(10n+9)}}} \qquad \text{(这里套Froullani公式)}\\
 &=\sum_{n=0}^{\infty}{\ln{\frac{(n+\frac{7}{10})(n+\frac{3}{10})}{(n+\frac{1}{10})(n+\frac{9}{10})}}}\\
 &=\lim_{n\rightarrow\infty}{\ln{\frac{\Gamma{(\frac{1}{10})}\Gamma{(\frac{9}{10})}\Gamma{(\frac{3}{10}+n)}\Gamma{(\frac{7}{10}+n)}}{\Gamma{(\frac{3}{10})}\Gamma{(\frac{7}{10})}\Gamma{(n+\frac{1}{10})}\Gamma{(n+\frac{9}{10})}}}}\qquad \text{(这里套Euler-Guess公式，也就是伽马函数的无穷乘积公式)}\\
 &=\ln{\frac{\sin{\frac{3\pi}{10}}}{\sin{\frac{\pi}{10}}}}\qquad \text{(这里套余元公式)}\\
 &=\ln{\frac{\sqrt{5}+1}{\sqrt{5}+1}}
\end{align*}

2.计算
\[ \int_{0}^{\infty}{\frac{e^{-x}(1-e^{-2x})}{x(e^{-4x}-e^{-2x}+1)(e^{-4x}+1)(e^{-8x}-e^{-4x}+1)}dx} \]
仿照上题的手段，我们有
\begin{align*}
 & \int_{0}^{\infty}{\frac{e^{-x}(1-e^{-2x})}{x(e^{-4x}-e^{-2x}+1)(e^{-4x}+1)(e^{-8x}-e^{-4x}+1)}dx}\\
 &=\int_{0}^{\infty}{\frac{e^{-x}(1-e^{-2x})(1+e^{-2x})(1-e^{-6x})}{x(e^{-4x}-e^{-2x}+1)(e^{-4x}+1)(e^{-8x}-e^{-4x}+1)(1+e^{-2x})(1-e^{-6x})} dx}\\
 &=\int_{0}^{\infty}{\frac{e^{-x}(1-e^{-4x})(1-e^{-6x})}{x(1-e^{-24x})}dx}\\
 &=\int_{0}^{\infty}{\sum_{n=0}^{\infty}{\frac{(1-e^{-4x})e^{-(24n+1)x}(1-e^{-6x})}{x}}dx}\\
 &=\int_{0}^{\infty}{\sum_{n=0}^{\infty}{ \frac{(e^{-(24n+1)x}-e^{-(24n+7)x})-(e^{-(24n+5)x}-e^{-(24n+11)x})}{x}}        dx}\\
 &=\sum_{n=0}^{\infty}{\ln{\frac{(24n+7)(24n+5)}{(24n+1)(24n+11)}}} \qquad \text{(这里套Froullani公式)}\\
 &=\sum_{n=0}^{\infty}{\ln{\frac{(n+\frac{7}{24})(n+\frac{5}{24})}{(n+\frac{1}{24})(n+\frac{11}{24})}}}\\
 &=\lim_{n\rightarrow\infty}{\ln{\frac{\Gamma{(\frac{1}{24})}\Gamma{(\frac{11}{24})}\Gamma{(\frac{7}{24}+n)}\Gamma{(\frac{5}{24}+n)}}{\Gamma{(\frac{7}{24})}\Gamma{(\frac{5}{24})}\Gamma{(n+\frac{1}{24})}\Gamma{(n+\frac{11}{24})}}}}\qquad \text{(这里套Euler-Guess公式，也就是伽马函数的无穷乘积公式)}\\
 &=\ln{\frac{\Gamma{(\frac{1}{24})}\Gamma{(\frac{11}{24})}}{\Gamma{(\frac{7}{24})}\Gamma{(\frac{5}{24})}}}
\end{align*}
另一方面
\begin{align*}
\frac{\Gamma{(\frac{1}{24})}\Gamma{(\frac{11}{24})}}{\Gamma{(\frac{5}{24})}\Gamma{(\frac{7}{24})}}
&=\frac{\Gamma{(\frac{1}{24})}\Gamma{(\frac{23}{24})}\Gamma{(\frac{11}{24})}\Gamma{(\frac{13}{24})}}{\Gamma{(\frac{5}{24})}\Gamma{(\frac{13}{24})}\Gamma{(\frac{7}{24})}\Gamma{(\frac{23}{24})}}\\
&=\frac{\pi^{2}}{\sin{(\frac{\pi}{24})}\cos{(\frac{\pi}{24})}}\cdot\frac{1}{\Gamma{(\frac{5}{24})}\Gamma{(\frac{13}{24})}\Gamma{(\frac{7}{24})}\Gamma{(\frac{23}{24})}}\\
&=\frac{2\pi^2}{\sin{(\frac{\pi}{12})}}\cdot\frac{\Gamma{(\frac{15}{24})}\Gamma{(\frac{21}{24})}}{\Gamma{(\frac{5}{24})}\Gamma{(\frac{13}{24})}\Gamma{(\frac{21}{24})}\Gamma{(\frac{7}{24})}\Gamma{(\frac{15}{24})}\Gamma{(\frac{23}{24})}}
\end{align*}
再利用公式
\[ \Gamma{(z)}\Gamma{\left(z+\frac{1}{3}\right)}\Gamma{\left(z+\frac{2}{3}\right)}=\frac{2\pi\sqrt{3}}{3^{2\pi}}\Gamma{(3z)} \]
我们有
\[ \Gamma{\left(\frac{5}{24}\right)}\Gamma{\left(\frac{13}{24}\right)}\Gamma{\left(\frac{21}{24}\right)}=\frac{2\pi\sqrt{3}}{3^{\frac{15}{24}}}\Gamma{\left(\frac{15}{24}\right)} \]
\[\Gamma{\left(\frac{7}{24}\right)}\Gamma{\left(\frac{15}{24}\right)}\Gamma{\left(\frac{23}{24}\right)}=\frac{2\pi\sqrt{3}}{3^{\frac{21}{24}}}\Gamma{\left(\frac{21}{24}\right)} \]
故
\[ \frac{\Gamma{(\frac{1}{24})}\Gamma{(\frac{11}{24})}}{\Gamma{(\frac{5}{24})}\Gamma{(\frac{7}{24})}}=\frac{\Gamma{(\frac{1}{24})}\Gamma{(\frac{23}{24})}\Gamma{(\frac{11}{24})}\Gamma{(\frac{13}{24})}}{\Gamma{(\frac{5}{24})}\Gamma{(\frac{13}{24})}\Gamma{(\frac{7}{24})}\Gamma{(\frac{23}{24})}}=\frac{3\sqrt{2}+\sqrt{6}}{2}
\]
所以
\[  \int_{0}^{\infty}{\frac{e^{-x}(1-e^{-2x})}{x(e^{-4x}-e^{-2x}+1)(e^{-4x}+1)(e^{-8x}-e^{-4x}+1)}dx}=\ln{\left(\frac{3\sqrt{2}+\sqrt{6}}{2} \right)} \]