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发表于 2020-7-25 11:38
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题:试证\(\small\,\displaystyle\sum_{m,n\ge 1}\frac{1}{mn(m+n)}=2\sum_{n=1}^{\infty}\frac{1}{n^3}=2\zeta(3).\)
证:\(\;\displaystyle{\small\,LHS=\sum_{m,n\ge 1}\int_0^1\int_0^1\int_0^1}(x^{m-1}y^{n-1}z^{m+n-1})dxdydz\)
\(\displaystyle\qquad\small\;=\int_{[0,1]^3}\big(z\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}(xz)^{m-1}(yz)^{n-1}\big)dV=\int_{[0,1]^3}\frac{z}{(1-xz)(1-yz)}dV\)
\(\displaystyle\qquad{\small\,=\int_0^1\frac{\ln^2(1-z)}{z}dz=\int_0^1\frac{\ln^2 x}{1-s}ds=\sum_{n=1}^{\infty}\int_0^1} s^{n-1}\ln^2 s ds\)
\(\displaystyle\qquad{\small\,=\sum_{n=1}^{\infty}\frac{2}{n^3}=\,}2\zeta(3).\quad\small\square\) |
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