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Playing around with double sums related to harmonic numbers and having easily found a closed expression for this sum

$$s_1 = \sum_{n,m\ge 1} \frac{1}{n m (n+m)}=2 \zeta(3)\tag{1}$$

I attempted to find a closed expression for this slightly more complicated sum

$$s_2 = \sum_{n,m\ge 1} \frac{1}{n m (n^2+m^2)}\tag{2}$$

which turned out to be more than I could achieve, and I hope for your help.

Here's what I did so far.

First approach

$$\begin{align}s_2 & = \sum_{n,m\ge 1} \frac{1}{n m (n^2+m^2)} \\ & = \sum_{n,m\ge 1} \frac{1}{n m }\int_{0}^{\infty } e^{-t(n^2+m^2)}\;dt\\ & =\int_{0}^{\infty }\left(\sum_{n,m\ge 1} \frac{1}{n m } e^{-t(n^2+m^2)}\right)\;dt\\ & =\int_{0}^{\infty } f_2(t)^2\;dt \end {align}\tag{3}$$

where

$$f_2(t) = \sum_{n\ge 1}\frac{e^{-t n^2}}{n}\tag{4}$$

Now

$$e^{-n^2 t}=\frac{1}{\sqrt{\pi}}\int_{-\infty }^{\infty } e^{-x^2+i 2 n x\sqrt{t} } \, dx\tag{5}$$

This linearizes $n$ in the exponent, and, exchanging integral and sum, we can do the (conditionally convergent) $n$-sum under the $x$-integral which gives

$$\frac{1}{\sqrt{\pi}}\sum _{n=1}^{\infty } \frac{e^{-x^2+i 2 n x \sqrt{t}}}{\sqrt{\pi } n}=-\frac{e^{-x^2}}{\sqrt{\pi}} \log \left(1-e^{2 i x \sqrt{t} }\right)\tag{6}$$

and

$$f_2(t) = -\frac{1}{\sqrt{\pi}}\int_{-\infty }^{\infty } e^{-x^2} \log \left(1-e^{2 i x \sqrt{t} }\right)\, dx\tag{7}$$

Now in the $x$-integral the imaginary part cancels out due to symmetry and the real part can be simplified so that

$$f_2(t) = -\frac{1}{\sqrt{\pi}}\int_{-\infty }^{\infty } e^{-x^2} \frac{1}{2} \log \left(4 \sin ^2\left(x\sqrt{t} \right)\right)\; dx\tag{8}$$

Alas, here I am stuck, and summing up it seems that I have made a simple looking formula more complicated.

Second approach

Doing just the $n$-sum in $s_2$ gives

$$s_2 = \sum_{m\ge 1} \frac{1}{2 m^3} (H(i m) + H(-i m))=\Re\left(\sum_{m\ge 1} \frac{1}{m^3} H(i m)\right)\tag{2.1}$$

where $H(z)$ is the harmonic number of argument $z$.

Inserting the representation

$$H(z) = \int_{0}^{1} \frac{1-x^z}{1-x}\;dx$$

and performing the $m$-sum gives an interesting integral representation of our double sum

$$s_2 = \left(\int_0^1 \frac{-\zeta (3)+\Re(\text{Li}_3\left(x^i\right))}{x-1} \, dx\right)\tag{2.2}$$

Jam
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Dr. Wolfgang Hintze
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  • (-1) What makes you think it has a particular closed form, to begin with? And what sort of closed form are you intending to aim for? The sum formula you've started off with is already a tidy expression, and unrestrictedly looking for anything else just seems like mindless rearrangements. – Jam Sep 28 '22 at 07:49
  • @ Jam I have shown a successful example of a closed form in the beginning of my question. Generally, I suggest you browse a little though the questions with the tag "harmonic numbers". You will find that this type of question is quite natural here, and you might reconsider the direction of your voting. – Dr. Wolfgang Hintze Sep 28 '22 at 08:04
  • @wolfgang It's an ill-defined goal to merely search for "a closed form", since it presents no clear end-point. If instead you were trying to establish that a number had a closed form as a rational (or as a sum or integral of elementary functions, or some such), then that would be a sensible goal since you could categorically say whether or not your number could. But otherwise, you're just aimlessly meandering in search of some alternative representation for the number, out of all possible representations and functions. (1/2) – Jam Sep 28 '22 at 08:14
  • I see no reason why we should expect a priori any sum to have a convenient closed form (in a specified language), given that even $\zeta(3)$ does not. And contrary to your point, most questions on the site with the [harmonic numbers] tag are well-posed. They are generally asking for proofs of con-/di-vergence, of equality, of derivations, and so forth. These have clear answers, unlike questions about closed forms. – Jam Sep 28 '22 at 08:19
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    @ Jam Sorry, but I cannot teach you the whole culture of closed form searches in this community. But you can do this by yourself as I have suggested earlier. – Dr. Wolfgang Hintze Sep 28 '22 at 08:23

1 Answers1

1

I am stuck almost at the same point.

Trying to get an asymptotics

$$S=\sum_{n= 1}^\infty \sum_{m= 1}^\infty\frac{1}{n m (n^2+m^2)}=\sum_{n= 1}^\infty \frac{H_{i n}+H_{-i n}}{2 n^3}$$

For "large" values of $n$ $$H_{i n}+H_{-i n}=2 (\log (n)+\gamma )+\frac{1}{6 n^2}+\frac{1}{60 n^4}+\frac{1}{126 n^6}+\frac{1}{120 n^8}+\frac{1}{66 n^{10}}+O\left(\frac{1}{n^{12}}\right)$$ is quite good as soon as $n \geq 5$

$$\left( \begin{array}{ccc} n & \text{exact} & \text{approximation} \\ 1 & 1.343731971 & 1.369186020 \\ 2 & 2.583614361 & 2.583605381 \\ 3 & 3.370392751 & 3.370392601 \\ 4 & 3.937503906 & 3.937503902 \\ 5 & 4.380001019 & 4.380001019 \end{array} \right)$$

Using the expansion for $n \geq 5$ does not make any problem leading to $ \color{red}{0.98978105099}37$ instead of $\color{red}{0.9897810509946}$

Claude Leibovici
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