Compute the double sum $\sum_{n, m>0, n \neq m} \frac{1}{n\left(m^{2}-n^{2}\right)}=\frac{3}{4} \zeta(3)$

Question

I am trying to compute the following double sum

$$\boxed{\sum_{n, m>0, n \neq m} \frac{1}{n\left(m^{2}-n^{2}\right)}=\frac{3}{4} \zeta(3)}$$

I proceeded as following

$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{n}\frac{1}{\left(m^2-n^2 \right)}=\sum_{n=1}^{\infty}\frac{1}{n}\sum_{m=1}^{\infty}\frac{1}{\left(m+n \right)\left(m-n \right)}$$

$$=\sum_{n=1}^{\infty}\frac{1}{n}\sum_{m=1}^{\infty}\frac{1}{2m}\left[\frac{1}{\left(m+n \right)}+\frac{1}{\left(m-n \right)}\right]$$

$$=\frac{1}{2}\underbrace{\sum_{n=1}^{\infty}\frac{1}{n}\sum_{m=1}^{\infty}\frac{1}{m}\frac{1}{\left(m+n \right)}}_{2\zeta(3)}+ \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n} \sum_{m=1}^{\infty}\frac{1}{m}\frac{1}{\left(m-n \right)}$$

The first sum is given here and I have also evaluated here, therefore we get that

$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{n}\frac{1}{\left(m^2-n^2 \right)}=\zeta(3)-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^2} \sum_{m=1}^{\infty}\frac{1}{m}-\frac{1}{\left(m-n \right)}$$

$$=\zeta(3)-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n^2} \int_{0}^{1}\frac{1-x^{-n}}{1-x}dx$$

$$=\zeta(3)-\frac{1}{2} \int_{0}^{1}\frac{1}{1-x}\left[ \sum_{n=1}^{\infty}\frac{1}{n^2}-\sum_{n=1}^{\infty}\frac{x^{-n}}{n^2}\right]dx$$

$$=\zeta(3)-\frac{1}{2} \int_{0}^{1}\frac{\zeta(2)-Li_{2}\left( \frac{1}{x} \right)}{1-x}dx$$

From this post we can use the relation

$$Li_{2}\left( \frac{1}{x} \right)-Li_{2}\left( x \right)=\zeta(2)-\frac{1}{2}\log^2(-x)$$

To get

$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{n}\frac{1}{\left(m^2-n^2 \right)}=\zeta(3)-\frac{\zeta(2)}{2} \int_{0}^{1}\frac{1}{1-x}dx+\frac{1}{2} \int_{0}^{1}\frac{Li_{2}\left( x \right)+\zeta(2)-\frac{1}{2}\log^2(-x)}{1-x}dx$$

$$=\zeta(3)+\frac{\zeta(2)}{2} \int_{0}^{1}\frac{1}{1-x}dx+\frac{1}{2} \int_{0}^{1}\frac{Li_{2}\left( x \right)}{1-x}dx-\frac{1}{4}\int_{0}^{1}\frac{\log^2(-x)}{1-x}dx$$

$$=\zeta(3)+\frac{\zeta(2)}{2} \log(1-x)\Big|_{0}^{1}+\frac{1}{2}\left\{ -Li_{2}(x)\log(1-x)\Big|_{0}^{1}-\int_{0}^{1}\frac{\log^2(1-x)}{x}dx\right\} -\frac{1}{4}\int_{0}^{1}\frac{\log^2(-x)}{1-x}dx$$

$$=\zeta(3)-\frac{1}{2}2\zeta(3) -\frac{1}{4}\int_{0}^{1}\frac{\log^2(-x)}{1-x}dx$$

$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{n}\frac{1}{\left(m^2-n^2 \right)}=-\frac{1}{4}\int_{0}^{1}\frac{\log^2(-x)}{1-x}dx$$

My questions are the following:

(1)Is the last equality correct?

(2)If so, what is the meaning of $\log^2(-x)=\log^2(e^{i\pi}x)$?

(3) How do we solve the last integral?

Answer 1

This was kind of tricky and the route I chose may not be justified, so I do expect to be chewed out.

Your sum is taken over $(m, n)$ in the first quadrant of $$C = \sum_{m \neq n} \frac{1}{n(m^2 - n^2)}$$

Start with partial fraction decomposition to get $$2C = \sum_{m \neq n } \Big(\frac{1}{n^2(m-n)} -\frac{1}{n^2(m+n)}\Big)$$ I will split the sum and "evaluate" the sums over each of the partial fractions.

Keep in mind the $(m, n)$ plane in the first quadrant, where m is the x-axis and n is the y-axis. Since $m \neq n$, we have that $$ m-n \neq 0$$ Consider each line $$ m - n = k$$ You will want to sum over these lines and adjust for n accordingly. When $k$ is positive, $n$ can be as small as one. When $k$ is negative, n must be at least $k+1$. Again, this can be seen if you keep the plane in mind. Since $m-n = k$:

$$\sum \frac{1}{n^2(m-n)} = \sum_{ k \geq 1, n \geq 1} \frac{1}{n^2k} +\sum_{ k \leq -1, n \geq k+1}\frac{1}{n^2k}$$ The sums on the RHS can be evaluated as $$\sum_{ k \geq 1, n \geq 1} \frac{1}{n^2k} +\sum_{ k \leq -1, n \geq k+1}\frac{1}{n^2k} = \zeta (2)\sum_{k \geq 1}\frac{1}{k} - \sum_{k \geq 1}\frac{(\zeta(2)-H^{2}_{k})}{k} = \sum_{k \geq 1}\frac{H^{2}_{k}}{k}$$ where $H^{2}_{k}$ is the harmonic sum $$H^{2}_{k} = \sum_{r \leq k} \frac{1}{r^2}$$ This is a divergent sum, but can be worked around.

In this sum we had to be careful about stepping on the points $(m, n)$ along the diagonal. In the second sum though, when we step on a diagonal, we get $$\sum_{m \geq 1} \frac{1}{m^2(m+m)}=\frac{\zeta(3)}{2}$$ so we can sum over the entire quadrant (except the x and y axes) and subtract this quantity from it to get $$\sum_{k \geq 1, 1 \leq n \leq k-1} \frac{1}{n^2(m+n)} = \sum_{k \geq 1, 1 \leq n \leq k-1} \frac{1}{n^2k}$$ where, keep in mind, the plane, and the fact that we are summing over lines $m+n = k$. This gives $$\sum_{k \geq 1, 1 \leq n \leq k-1} \frac{1}{n^2k} = \sum_{k \geq 1}\frac{\big(1 +\frac{1}{2^2} + ... + \frac{1}{(k-1)^2}\big)}{k} = \sum_{ k \geq 1} \frac{H^{2}_{k-1}}{k} = \sum_{k \geq 1}\frac{H^2_{k} - \frac{1}{k^2}}{k} = -\zeta(3)+\sum_{k \geq 1}\frac{H^2_{k}}{k}$$ To find $2C$, we return to the original expression and plug things back in: $$2C = \sum_{m \neq n } \Big(\frac{1}{n^2(m-n)} -\frac{1}{n^2(m+n)}\Big) = \sum_{k \geq 1}\frac{H^2_{k}}{k} - \big(-\zeta(3) - \frac{\zeta(3)}{2} +\sum_{k \geq 1} \frac{H^2_{k}}{k} \big) = \frac{3\zeta(3)}{2}$$ Therefore, $$C = \frac{3\zeta(3)}{4}$$

Answer 2

In here, we have that $\displaystyle \sum_{\underset{m \neq n}{m = 1}}^{\infty} \frac{1}{m^2 - n^2} = \frac{3}{4n^2}$, therefore

\begin{aligned} \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{n}\frac{1}{\left(m^2-n^2 \right)} = \frac{3}{4}\sum_{n=1}^{\infty} \frac{1}{n^3} = \frac{3}{4}\zeta(3).\end{aligned}

Answer 3

This doesn't really answer your questions, however consider another approach perhaps as a way of side-lining somewhat difficult integrals

$$\sum\limits_{n,m>0,n\ne m}{\frac{1}{n\left( {{m}^{2}}-{{n}^{2}} \right)}}=\frac{1}{2}\sum\limits_{n'=1}^{\infty }{\sum\limits_{m'=1}^{\infty }{\frac{1}{{{n}^{2}}\left( m-n \right)}}}-\frac{1}{2}\sum\limits_{n=1}^{\infty }{\sum\limits_{m=1}^{\infty }{\frac{1}{{{n}^{2}}\left( m+n \right)}}}+\frac{1}{4}\sum\limits_{m=1}^{\infty }{\frac{1}{{{m}^{3}}}}$$ Primes on sums denote excluding $n=m$. Note $$\psi \left( -z \right)+\gamma \underset{z\to n}{\mathop{=}}\,\frac{1}{z-n}+{{H}_{n}}+\sum\limits_{k=1}^{\infty }{\left( {{\left( -1 \right)}^{k}}H_{n}^{k+1}-\zeta \left( k+1 \right) \right){{\left( z-n \right)}^{k}}},n\ge 0$$ So we have by the theorem that the sum over all residues is zero $$\sum\limits_{n,m>0,n\ne m}{\frac{1}{n\left( {{m}^{2}}-{{n}^{2}} \right)}}=\frac{1}{2}\sum\limits_{m=1}^{\infty }{\left\{ -\underset{z=0}{\mathop{res}}\,\frac{\psi \left( -z \right)+\gamma }{{{z}^{2}}\left( m-z \right)}+\underset{z=-m,0}{\mathop{res}}\,\frac{\psi \left( -z \right)+\gamma }{{{z}^{2}}\left( m+z \right)}+\frac{1}{2{{m}^{3}}} \right\}}$$ The first residue calculation specifically excludes the residue at $z=m$ to avoid the singularities at $n=m$. Using the expansion above to determine the residues we find $$\sum\limits_{n,m>0,n\ne m}{\frac{1}{n\left( {{m}^{2}}-{{n}^{2}} \right)}}\\=\frac{1}{2}\sum\limits_{m=1}^{\infty }{\left\{ \frac{{{m}^{2}}{{\pi }^{2}}-6}{6{{m}^{3}}}+\frac{-1+\gamma +\psi \left( m \right)}{{{m}^{2}}}+\frac{6+6m-{{m}^{2}}{{\pi }^{2}}}{6{{m}^{3}}}+\frac{1}{2{{m}^{3}}} \right\}}$$ Cleaning this up $$\sum\limits_{n,m>0,n\ne m}{\frac{1}{n\left( {{m}^{2}}-{{n}^{2}} \right)}}=\frac{1}{2}\sum\limits_{m=1}^{\infty }{\left\{ \frac{\gamma +\psi \left( m \right)}{{{m}^{2}}}+\frac{1}{2{{m}^{3}}} \right\}}=\frac{1}{2}\left\{ \zeta \left( 3 \right)+\frac{1}{2}\zeta \left( 3 \right) \right\}=\frac{3}{4}\zeta \left( 3 \right)$$ Where we’ve used the reasonably well known representation $$\zeta \left( 3 \right)=\sum\limits_{m=1}^{\infty }{\frac{\gamma +\psi \left( m \right)}{{{m}^{2}}}}$$