15
$\begingroup$

I'm trying to determine closed forms for any of the sums in the following expressions using either harmonic summation techniques, the residue theorem, Abel-Plana Formula or Euler-Maclaurin Formula:

It is pretty well known that $\zeta (3)$ can be expressed as any of the following: $$\zeta (3) = \frac {7 \pi^3} {180} - 2\sum_{k=1}^{\infty} \frac {1} {k^3 (e^{2 \pi k} -1)}$$ $$\zeta (3) = \frac {\pi^3} {28} + \frac{16}{7}\sum_{k=1}^{\infty} \frac {1} {k^3 (e^{ \pi k} +1)} - \frac{2}{7} \sum_{k=1}^{\infty} \frac {1} {k^3 (e^{2 \pi k} +1)}$$ $$\zeta (3) = \frac {67 \pi^3} {1620} - \frac{16}{9}\sum_{k=1}^{\infty} \frac {1} {k^3 (e^{ \pi k} -1)} + \frac{2}{9} \sum_{k=1}^{\infty} \frac {1} {k^3 (e^{2 \pi k} +1)}$$ $$\zeta (3) = \frac {37\pi^3} {900} - \frac{2}{5}\sum_{k=1}^{\infty} \frac {4} {k^3 (e^{ \pi k} -1)} - \frac{2}{5} \sum_{k=1}^{\infty} \frac {1} {k^3 (e^{4 \pi k} -1)}$$ $$\zeta (3) = \frac {\sqrt {2} \pi^3} {36} - \frac{4}{3}\sum_{k=1}^{\infty} \frac {1} {k^3 (e^{ \pi k \sqrt{2}} -1)} - \frac{2}{3} \sum_{k=1}^{\infty} \frac {1} {k^3 (e^{2 \pi k \sqrt{2}} -1)}$$ By applying a difference of squares to the first sum, the following can be obtained: $$\zeta (3) = \frac {7 \pi^3} {180} - \sum_{k=1}^{\infty} \frac {1} {k^3 (e^{\pi k} - 1)} + \sum_{k=1}^{\infty} \frac {1} {k^3 (e^{\pi k} + 1)} $$ From here, I was able to deduce closed forms for sums involving hyperbolic trigonometric functions, with the use of Euler's Formula $e^{i \theta} = \cos (\theta) + i \sin (\theta)$, such as: $$\sum_{k=1}^{\infty} \frac {\coth \left( \frac {\pi k} {2} \right) + \tanh \left( \frac {\pi k} {2} \right)} {k^3} = \frac {7 \pi^3} {90}$$ $$\sum_{k=1}^{\infty} \frac {8\coth \left( \frac {\pi k} {2} \right) + \tanh \left( \pi k \right)} {k^3} = \frac {67 \pi^3} {180}$$ $$\sum_{k=1}^{\infty} \frac {8 \coth \left( \frac {\pi k} {2} \right) + \coth ( \pi k) + \tanh (\pi k)} {k^3} = \frac {37 \pi^3} {90}$$ $$\sum_{k=1}^{\infty} \frac {4 \coth \left( \frac {\pi k} {2} \right) - 4 \tanh \left( \frac {\pi k} {2} \right) + \tanh ( \pi k)} {k^3} = \frac {11 \pi^3} {180}$$ And my personal favourite, $$\sum_{k=1}^{\infty} \frac {8\tanh \left( \frac {\pi k} {2} \right) - \tanh \left( \pi k \right)} {k^3} = \frac {\pi^3} {4}$$

Additional closed forms can be obtained by repeated operations: $$\sum_{k=1}^{\infty} \frac {-2 \coth \left( \frac {\pi k} {2} \right) + 6 \tanh \left( \frac {\pi k} {2} \right) - \tanh ( \pi k)} {k^3} = \frac {17 \pi^3} {180}$$ $$\sum_{k=1}^{\infty} \frac { -\coth \left( \frac {\pi k} {2} \right) + 7 \tanh \left( \frac {\pi k} {2} \right) - \tanh ( \pi k)} {k^3} = \frac {31 \pi^3} {180}$$ $$\sum_{k=1}^{\infty} \frac {-3 \coth \left( \frac {\pi k} {2} \right) + 5 \tanh \left( \frac {\pi k} {2} \right) - \tanh ( \pi k)} {k^3} = \frac {\pi^3} {60}$$ $$\sum_{k=1}^{\infty} \frac {7 \coth \left( \frac {\pi k} {2} \right) - \tanh \left( \frac {\pi k} {2} \right) + \tanh ( \pi k)} {k^3} = \frac {53 \pi^3} {180}$$ $$\sum_{k=1}^{\infty} \frac {7 \coth \left( \frac {\pi k} {2} \right) + \coth (\pi k) - \tanh \left( \frac {\pi k} {2} \right) + \tanh ( \pi k)} {k^3} = \frac {\pi^3} {3}$$

etc.

Even with this information, I found the sums would not assist me in finding $\zeta (3)$ due to it being cancelled when separating the original sums into hyperbolic trigonometric forms. However, potentially the following expression may show some use for hyperbolic trigonometry for evaluating $\zeta (3)$: $$\zeta (3) = \lim_{\alpha \to 0+} \frac {\pi^2} {7} \sum_{k=1}^{\infty} \frac {\sinh (k \alpha )} {k \cosh (k \alpha )^3}$$

I then decided to turn my attention towards the original difference of squares equation. When I tried to look into how to potentially evaluate closed forms to sums that appear like this, the following question assisted me, as well as this question, and this one.

The top response uses the Mellin Transform to evaluate the sum, and I believe that the same could potentially be done for the sum that I am interested in. Applying the Mellin Transform to $\sum_{k=1}^{\infty} \frac {1} {k^3 (e^{\pi k} + 1)}$ : $$F(s) = \int_{0}^{\infty} x^{s-1} \left( \frac {1}{x^3 \left( e^{\pi x} + 1 \right)}\right) dx$$ $$F(s) = 2^{-s} (2^{s} - 16) \pi^{3-s} \zeta (s-3) \Gamma (s-3)$$ Applying the inverse Mellin Transform (Note that I'm not entirely sure what the following contour is and this is essentially my question, as well as how to calculate residue; the top answer in the question linked above used $\int_{c - i \infty}^{c + i \infty}$): $$\frac {1} {x^3 (e^{\pi x} + 1)} = \frac {1} {2 \pi i} \int_C 2^{-s} (2^s - 16) \pi^{3-s} \zeta (s-3) \Gamma (s-3) x^{-s} ds$$ Substituting $x=k$, and summing the terms: $$\sum_{k=1}^{\infty} \frac {1} {k^3 (e^{\pi k} + 1)} = \frac {1} {2 \pi i} \int_C 2^{-s} (2^s - 16) \pi^{3-s} \zeta (s-3) \Gamma (s-3) \sum_{k=1}^{\infty} k^{-s} ds$$ $$\sum_{k=1}^{\infty} \frac {1} {k^3 (e^{\pi k} + 1)} = \frac {1} {2 \pi i} \int_C 2^{-s} (2^s - 16) \pi^{3-s} \zeta (s-3) \Gamma (s-3) \zeta (s) ds$$

This now brings me onto my question; am I making a misconception or could the Mellin Transform be used here? Additionally, how does one calculate the residue of this integral? What happens in the case of simple poles versus higher-order poles? If this can't be used here, could it instead be used in any other of the other corresponding sums at the start of my post or via the use of a different contour shape that could make at least one of these possible to calculate a closed-form for? Such as:

Similarly, the Mellin Transform applied to $\sum_{k=1}^{\infty} \frac {1} {k^3 (e^{\pi k} - 1)}$ : $$\sum_{k=1}^{\infty} \frac {1} {k^3 (e^{\pi k} - 1)} = \frac {1} {2 \pi i} \int_C \pi^{3-s} \mathrm{Li}_{s-3} (1) \Gamma (s-3) \zeta (s) ds$$ Where $\mathrm{Li}$ represents the Polylogarithm Function.

$\endgroup$
14
  • 6
    $\begingroup$ If there were any doubt: the elusiveness of $\zeta(3)$ itself is not surprising, given its history, and all the things that people have attempted... $\endgroup$ – paul garrett Feb 9 at 23:12
  • 2
    $\begingroup$ Is $(2\pi)^{-s}\zeta(s)\zeta(s-3)\Gamma(s)$ and related functions decreasing fast enough as $\Re(s)\to -\infty$ to apply the residue theorem? $\endgroup$ – reuns Feb 10 at 14:44
  • 1
    $\begingroup$ This is also an issue in the linked answer. However, the integral appearing there has a symmetry w.r.t. $s\mapsto-s$, which allows to use a "tall rectangular" contour there. In our case, there doesn't seem to be such a symmetry ($s\mapsto 4-s$ isn't quite good). $\endgroup$ – metamorphy Feb 10 at 17:49
  • 2
    $\begingroup$ Found an article following this approach. Looks as good as one gets. $\endgroup$ – metamorphy Feb 10 at 19:21
  • 1
    $\begingroup$ The closest thing to a closed-form expression for $\zeta(3)$ that I've seen is $$ \zeta(3) = 4 \pi^2 \log B,$$ where $B$ is the Glaisher-Kinkelin-like constant $$ \log B = \lim_{n \to \infty} \left[\sum_{k=1}^{n} k^{2} \log k- \left(\frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6} \right) \log n + \frac{n^{3}}{9} - \frac{n}{12} \right].$$ $\endgroup$ – Random Variable Feb 17 at 13:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.