A Beautiful Double Sum from @infseriesbot

I saw this beautiful result below and gave it a try, here is my solution:

$S=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{nm(n+m)}$

After partial fraction decomposition

$S=\sum_{m=1}^{\infty}\frac{1}{m^2}\underbrace{\sum_{n=1}^{\infty}\frac{1}{n}-\frac{1}{m+n}}_{=\psi(m+1)+\gamma}$

Recall Digamma´s integral representation

$\boxed{\psi(m+1)+\gamma=\int_{0}^{1}\frac{1-t^m}{1-t}dt}$

We get

$S=\sum_{m=1}^{\infty}\frac{1}{m^2}\int_{0}^{1}\frac{1-t^m}{1-t}dt$

$=\int_{0}^{1}\frac{1}{1-t}\sum_{m=1}^{\infty}\frac{1-t^m}{m^2}dt$

$=\int_{0}^{1}\frac{1}{1-t}\Big\{\sum_{m=1}^{\infty}\frac{1}{m^2}-\sum_{m=1}^{\infty}\frac{t^m}{m^2} \Big \}dt$

$=\int_{0}^{1}\frac{1}{1-t}\Big\{\zeta(2)-Li_{2}(t) \Big \}dt$

$=\zeta(2)\int_{0}^{1}\frac{1}{1-t}dt-\int_{0}^{1}\frac{Li_{2}(t)}{1-t}dt$

$=-\zeta(2)\lim_{t \rightarrow 1}\ln(1-t)-\int_{0}^{1}\frac{Li_{2}(t)}{1-t}dt$

Now, recall integral representation of the Dilogarithm function

$\boxed{Li_{2}(t)=-\int_{0}^{t}\frac{\ln(1-y)}{y}dy}$

$S=-\zeta(2)\lim_{t \rightarrow 1}\ln(1-t)+\int_{0}^{1}\frac{1}{1-t}\int_{0}^{t}\frac{\ln(1-y)}{y}dydt$

Swapping order of integration

$S=-\zeta(2)\lim_{t \rightarrow 1}\ln(1-t)+\int_{0}^{1}\frac{\ln(1-y)}{y} \Big(\int_{y}^{1}\frac{dt}{1-t} \Big)dy$

$=-\zeta(2)\lim_{t \rightarrow 1}\ln(1-t)+\int_{0}^{1}\frac{\ln(1-y)}{y} \Big(-\lim_{t \rightarrow 1}\ln(1-t)+\log(1-y)\Big)dy$

$=-\zeta(2)\lim_{t \rightarrow 1}\ln(1-t)-\lim_{t \rightarrow 1}\ln(1-t)\int_{0}^{1}\frac{\ln(1-y)}{y} dy+ \int_{0}^{1}\frac{\ln^2(1-y)}{y} dy$

$=-\zeta(2)\lim_{t \rightarrow 1}\ln(1-t)+\lim_{t \rightarrow 1}\ln(1-t)\int_{0}^{1}\frac{1}{y}\sum_{k=1}^{\infty}\frac{y^k}{k} dy+ \int_{0}^{1}\frac{\ln^2(1-y)}{y} dy$

$=-\zeta(2)\lim_{t \rightarrow 1}\ln(1-t)+\lim_{t \rightarrow 1}\ln(1-t)\sum_{k=1}^{\infty}\frac{1}{k}\int_{0}^{1}y^{k-1} dy+ \int_{0}^{1}\frac{\ln^2(1-y)}{y} dy$

$=-\zeta(2)\lim_{t \rightarrow 1}\ln(1-t)+\zeta(2)\lim_{t \rightarrow 1}\ln(1-t)+ \int_{0}^{1}\frac{\ln^2(1-y)}{y} dy$

$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{nm(n+m)}=\int_{0}^{1}\frac{\ln^2(1-y)}{y} dy$

Change variable 1-y=t

$=\int_{0}^{1}\frac{\ln^2(t)}{1-t} dt$

$=\int_{0}^{1}\ln^2(t)\sum_{k=0}^{\infty}t^kdt$

$=\sum_{k=0}^{\infty}\int_{0}^{1}t^k\ln^2(t)dt$

Now use the fact that

$\boxed{\int_{0}^{1}x^m ln^n(x)dx=\frac{(-1)^{n}n! }{(m+1)^{n+1}}}$

for $n=2 \,\,\text{and} \,\, m=k$

$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{nm(n+m)}=\sum_{k=0}^{\infty}\frac{(-1)^{2}2! }{(k+1)^{3}}$

$\boxed{\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{nm(n+m)}=2\zeta(3)}$

Corollary

We have shown right in the beginning of the proof that

$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{nm(n+m)}=\sum_{m=1}^{\infty}\frac{1}{m^2}\int_{0}^{1}\frac{1-t^m}{1-t}dt$

The integral on the RHS of the equation above is an integral representation of $H_{m}$ , the harmonic number.Therefore we may right

$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{nm(n+m)}=\sum_{m=1}^{\infty}\frac{H_{m}}{m^2}$

and we conclude that

$\boxed{\sum_{m=1}^{\infty}\frac{H_{m}}{m^2}=2\zeta(3)}$

Apendix

Proof that

$\psi(m+1)=-\gamma+\int_{0}^{1}\frac{1-t^m}{1-t}dt$

Recall the series representation of the digamma function derived from the Weierstrass infinite product of the gamma function

$\psi(m+1) &=-\gamma+\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+m}\right)$

$=-\gamma+\sum_{n=1}^{\infty} \int_{0}^{1}\left(x^{n-1}-x^{n+m-1}\right) d x$

$=-\gamma+\int_{0}^{1} \sum_{n=1}^{\infty}\left(x^{n-1}-x^{n+m-1}\right) d x$

$=-\gamma+\int_{0}^{1} \sum_{n=1}^{\infty}x^{n-1}\left(1-x^{m}\right) d x$

$=-\gamma+\int_{0}^{1} \left(1-x^{m}\right)\sum_{n=1}^{\infty}x^{n-1} d x$

$\boxed{\psi(m+1) =-\gamma+\int_{0}^{1} \frac{1-x^{m}}{1-x} d x}$

Proof that

$Li_{2}(t)=-\int_{0}^{t}\frac{\ln(1-y)}{y}dy$

$=-\int_{0}^{t}\frac{1}{y} \Big( -\sum_{k=1}^{\infty}\frac{y^k}{k}\Big)dy$

$=\sum_{k=1}^{\infty}\frac{1}{k}\int_{0}^{t} y^{k-1}dy$

$=\sum_{k=1}^{\infty}\frac{1}{k} \Big[\frac{t^k}{k} \Big]$

$\boxed{-\int_{0}^{t}\frac{\ln(1-y)}{y}dy=\sum_{k=1}^{\infty}\frac{t^k}{k^2}}$

Proof that

$\int_{0}^{1}x^m ln^n(x)dx=\frac{(-1)^{n}n! }{(m+1)^{n+1}}$

Let $x=e^{-y} \Rightarrow dx= -e^{-y}dy$ , the integral becomes

$\int_{0}^{\infty}e^{-my}\ln^{n}(e^{-y})e^{-y}dy$

$=\int_{0}^{\infty}e^{-(m+1)y}(-1)^{n}y^{n})dy$

Now let (m+1)y=u then

$\int_{0}^{1}x^m ln^n(x)dx=\frac{(-1)^{n}}{(m+1)^{n+1}}\int_{0}^{\infty}e^{-u}u^ndu$

Finally!

$\boxed{\int_{0}^{1}x^m ln^n(x)dx=\frac{(-1)^{n}n!}{(m+1)^{n+1}}}$

Ricardo Albahari

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