试证：∑(i=1,∞)∑(j=1,∞)1／[ij(i+j)]=2ζ(3)

elim

题：试证 \(\displaystyle\sum_{i,j=1}^{+\infty} \frac{1}{ij(i+j)}=2\zeta(3).\)

elim

\(\displaystyle\sum_{i,j=1}^{+\infty} \frac{1}{ij(i+j)} = \sum_{i,j=1}^{+\infty} \int_{0}^1 \frac{x^{i+j-1}}{ij}  \mathrm{d}x= \int_{0}^{1}\left(\sum_{i,j=1}^{+\infty}   \frac{x^{i+j}}{ij}\right) \frac{\mathrm{d}x}{x}\)
\( \displaystyle=  \int_{0}^{1} \frac{\ln^2(1-x)}{x} \mathrm{d}x = \int_{0}^{1} \frac{\ln^2(x)}{1-x} \mathrm{d}x =  \int_{0}^{1}\sum_{k=0}^{\infty}x^k \ln^2 (x)\mathrm{d}x\)
\(\displaystyle=\sum_{k=0}^{\infty}  \int_{0}^{1}x^k \ln^2 (x)\mathrm{d}x  = \sum_{k=0}^{\infty} \frac{2}{(k+1)^3} = 2 \zeta (3)\)

luyuanhong

楼上 elim 的帖子很好！已收藏。