阅读此贴,受益匪浅。这是几个从此贴可导出的结果(4阶),5阶的例子我发在楼主另外一贴了
![](http://tb2.bdstatic.com/tb/editor/images/face/i_f24.png?t=20140803)
$$\int_0^1 \frac{\text{Li}_4(x)}{(x (1-x))^{3/4}} \, dx=\frac{\Gamma \left(\frac{1}{4}\right)^2}{2 \sqrt{\pi }} \left(\frac{5 \pi ^2 C}{3}-4 C \log ^2(2)-4 \pi C \log (2)+32 \Im(\text{Li}_4(1+i))-22 \text{Li}_4\left(\frac{1}{2}\right)-7 \pi \zeta (3)-14 \zeta (3) \log (2)+\frac{277 \pi ^4}{960}-\log ^4(2)-\frac{1}{2} \pi \log ^3(2)+\frac{9}{8} \pi ^2 \log ^2(2)-\frac{3}{8} \pi ^3 \log (2)-\frac{\psi ^{(3)}\left(\frac{1}{4}\right)}{96}+\frac{\psi ^{(3)}\left(\frac{3}{4}\right)}{96}\right)$$
$$\int_0^1 \frac{\text{Li}_4\left(\sqrt{x}\right)}{(x (1-x))^{3/4}} \, dx=\frac{\Gamma \left(\frac{1}{4}\right)^2 }{2 \sqrt{\pi }}\left(\frac{17 \pi ^2 C}{48}-\frac{1}{4} C \log ^2(2)-\frac{3}{4} \pi C \log (2)+2 \Im(\text{Li}_4(1+i))-\frac{11 \text{Li}_4\left(\frac{1}{2}\right)}{8}-\frac{21 \pi \zeta (3)}{16}-\frac{7}{8} \zeta (3) \log (2)+\frac{239 \pi ^4}{5120}-\frac{1}{16} \log ^4(2)-\frac{5}{96} \pi \log ^3(2)+\frac{13}{128} \pi ^2 \log ^2(2)-\frac{1}{128} \pi ^3 \log (2)+\frac{\psi ^{(3)}\left(\frac{3}{4}\right)}{1536}-\frac{\psi ^{(3)}\left(\frac{1}{4}\right)}{1536}\right)$$
$$\int_0^1 \frac{\text{Li}_4(x)}{\sqrt[4]{x (1-x)}} \, dx=\frac{\Gamma \left(\frac{3}{4}\right)^2}{\sqrt{\pi }} \left(-\frac{5 \pi ^2 C}{3}-8 \pi C+4 C \log ^2(2)-4 \pi C \log (2)+16 C \log (2)-32 \Im(\text{Li}_3(1+i))-32 \Im(\text{Li}_4(1+i))-22 \text{Li}_4\left(\frac{1}{2}\right)+\frac{21 \zeta (3)}{2}+7 \pi \zeta (3)-14 \zeta (3) \log (2)+\frac{277 \pi ^4}{960}+\frac{3 \pi ^3}{4}+\frac{5 \pi ^2}{3}+8 \pi -32-\log ^4(2)-\frac{2 \log ^3(2)}{3}+\frac{1}{2} \pi \log ^3(2)+\frac{9}{8} \pi ^2 \log ^2(2)+3 \pi \log ^2(2)-4 \log ^2(2)+\frac{3}{8} \pi ^3 \log (2)+\frac{5}{6} \pi ^2 \log (2)+4 \pi \log (2)-16 \log (2)+\frac{\psi ^{(3)}\left(\frac{1}{4}\right)}{96}-\frac{\psi ^{(3)}\left(\frac{3}{4}\right)}{96}\right)$$
$$\int_0^1 \frac{\text{Li}_4\left(\sqrt{x}\right)}{\sqrt[4]{x (1-x)}} \, dx=-\sqrt{2} \pi +\frac{\Gamma \left(\frac{3}{4}\right)^2 }{\sqrt{\pi }}\left(-\frac{3 \pi C}{2}-\frac{17 \pi ^2 C}{48}+\frac{1}{4} C \log ^2(2)-\frac{3}{4} \pi C \log (2)+C \log (2)-2 \Im(\text{Li}_3(1+i))-2 \Im(\text{Li}_4(1+i))-\frac{11 \text{Li}_4\left(\frac{1}{2}\right)}{8}+\frac{21 \zeta (3)}{32}+\frac{21 \pi \zeta (3)}{16}-\frac{7}{8} \zeta (3) \log (2)+\frac{239 \pi ^4}{5120}+\frac{\pi ^3}{64}+\frac{17 \pi ^2}{48}+\frac{3 \pi }{2}-2-\frac{1}{16} \log ^4(2)-\frac{\log ^3(2)}{24}+\frac{5}{96} \pi \log ^3(2)+\frac{13}{128} \pi ^2 \log ^2(2)-\frac{\log ^2(2)}{4}+\frac{5}{16} \pi \log ^2(2)+\frac{1}{128} \pi ^3 \log (2)+\frac{17}{96} \pi ^2 \log (2)+\frac{3}{4} \pi \log (2)-\log (2)+\frac{\psi ^{(3)}\left(\frac{1}{4}\right)}{1536}-\frac{\psi ^{(3)}\left(\frac{3}{4}\right)}{1536}\right)$$