Feynman's trick V

Double sum and Feynman's trick

We prove the following double sum with the help of the Feynman's trick \[\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{nm(n+m)} = 2\zeta(3)\] We also found the following generalization \[\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{nm(rn+m)} = \frac{1}{r}\sum_{k=1}^{\infty} \frac{H_{rk}}{k^2}\] Proof Recall \[ \int_{0}^{1} w^{n-1} dw = \frac{1}{n} \] \[ \int_{0}^{1} y^{n-1} dy = \frac{1}{m}\] \[ \int_{0}^{1} x^{n+m-1} dx = \frac{1}{n+m}\] Therefore \[\frac{1}{nm(n+m)} = \int_{0}^{1}\int_{0}^{1}\int_{0}^{1} w^{n-1}y^{m-1}x^{n+m-1}dwdydx\] Summing \begin{align*} \sum_{n=1}^{\infty}\frac{1}{nm(n+m)} =& \int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \sum_{n=1}^{\infty} w^{n-1}y^{m-1}x^{n+m-1}dwdydx\\ & = \int_{0}^{1}\int_{0}^{1}\int_{0}^{1} \frac{y^{m-1} x^{m}}{(1-wx)}dwdydx \end{align*} Then \begin{align*} \sum_{m=1}^{\infty} \sum_{n=1}^{\infty}\frac{1}{nm(n+m)} =& \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \sum_{m=1}^{\infty} \frac{y^{m-1} x^{m}}{(1-wx)}dwdydx\\ =& \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{x}{(1-wx)(1-yx)}dwdydx\\ =& \int_{0}^{1} \int_{0}^{1} \frac{x}{1-yx} \left[\int_{0}^{1} \frac{1}{(1-wx)}dw\right]dydx\\ =& \int_{0}^{1} \left[\int_{0}^{1} \frac{\ln(1-x)}{1-yx} dy\right]dx\\ =& \int_{0}^{1} \frac{\ln^2(1-x)}{x}dx\\ =& \int_{0}^{1} \frac{\frac{d^2}{dt^2}\Big|_{t=0+} (1-x)^t}{x}dx\\ =& \frac{d^2}{dt^2}\Big|_{t=0+} \int_{0}^{1} \frac{(1-x)^t}{x}dx\\ =& \frac{d^2}{dt^2}\Big|_{t=0+} \int_{0}^{1}\sum_{n=0}^{\infty} (1-x)^n(1-x)^tdx \\ =& \frac{d^2}{dt^2}\Big|_{t=0+} \sum_{n=0}^{\infty}\int_{0}^{1} (1-x)^{n+t}dx \\ =& \lim_{t \to 0+ }\frac{d^2}{dt^2} \sum_{n=0}^{\infty} \frac{1}{n+t+1}\\ =& \lim_{t \to 0+ } \sum_{n=0}^{\infty} \frac{2}{(n+t+1)^3}\\ =& \sum_{n=0}^{\infty} \frac{2}{(n+1)^3}\\ =& 2\zeta(3) \end{align*} \[\boxed{\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{nm(n+m)} = 2\zeta(3)}\] For the general case, \begin{align*} \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{nm(rn+m)} =& \int_{0}^{1}\frac{ \ln(1-x)\ln(1-x^r)}{x} dx\\ = & \int_{0}^{1}\frac{ \frac{d}{dt}\Big|_{t=0+} (1-x)^t \ln(1-x^r)}{x} dx\\ = & \frac{d}{dt}\Big|_{t=0+} \int_{0}^{1}\frac{ (1-x)^t \ln(1-x^r)}{x} dx\\ = & -\frac{d}{dt}\Big|_{t=0+} \int_{0}^{1} \sum_{k=1}^{\infty}\frac{1}{k} (1-x)^t x^{rk-1} dx \\ = & -\frac{d}{dt}\Big|_{t=0+} \sum_{k=1}^{\infty} \frac{1}{k} \int_{0}^{1} (1-x)^t x^{rk-1} dx \\ = & -\frac{d}{dt}\Big|_{t=0+} \sum_{k=1}^{\infty}\frac{1}{k} B(rk,t+1) \\ =& \lim_{t \to 0+} -\sum_{k=1}^{\infty}\frac{\left(\psi^{0}(t+1)-\psi^{0} (kr+t+1)\right)B(kr,t+1)}{k}\\ =& \sum_{k=1}^{\infty}\frac{\left(\psi^{0}(kr+1) + \gamma \right)}{k^2r}\\ =& \frac{1}{r} \sum_{k=1}^{\infty}\frac{H_{kr}}{k^2} \end{align*} \[\boxed{\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{nm(rn+m)} = \frac{1}{r}\sum_{k=1}^{\infty} \frac{H_{rk}}{k^2}}\]