The series $$ S_{1/2} := \sum_{k\in\mathbb Z} \frac{(-1)^k}{(2k+1)^3} $$ can be summed using a rather standard trick, namely by defining a suitable meromorphic function $$ f(z) := \frac{\csc(\pi z)}{(2z + 1)^3} $$ and observing that its contour integral on an infinite-radius circle around the origin evaluates to both 0 (by asymptotic analysis) and $ 2\pi \mathrm{i}\left(S_{1/2}/\pi - \pi^2/16\right) $ (by the residue theorem), which entails $ S_{1/2} = \pi^3/16 $.

I am trying to use the same approach to compute a generalized version of that series, $$ S_\alpha := \sum_{k\in\mathbb Z} \frac{\sin\left[(2k + 1)\pi\alpha\right]}{(2k+1)^3}, $$ which, I have reasons to believe, should be equal to $ S_\alpha = \pi^3\alpha (1-\alpha)/4 $.

I can now proceed in a similar way as before and note that $$ f_\alpha(z) := \frac{\sin\left[(2z + 1)\pi\alpha\right] \cot(\pi z)}{(2z + 1)^3} $$ has the property that $$ \lim_{n\to\infty} \frac{1}{2\pi\mathrm{i}} \oint_{C_n} f_\alpha(z)\,\mathrm{d}z = \frac{S_\alpha}{\pi} + \frac{\pi^2\alpha}{8}, $$ where $ {\{C_n\}}_n $ is a sequence of positive-oriented circles with radius going to infinity without ever intercepting any poles. The problem is that the left hand side is not 0 anymore (because the integrand is now unbounded in the complex plane), so this does not give any useful information unless I am able to also compute the residue at infinity of $ f_\alpha $. However, calculating the residue of $ f_\alpha(1/z)/z^2 $ is hard, because the function develops an essential singularity at $ 1/z = 0 $, and the residue itself ends up being expressed as a nasty series—much nastier-looking, in fact, than the one I was trying to compute in the first place.

I would appreciate any help, whether a simple hint or a full solution, to crack this series. Thank you very much in advance.

  • $\begingroup$ Are you interested in other ways to evaluate your series? $\endgroup$ – Simply Beautiful Art Jan 12 '17 at 12:22
  • $\begingroup$ @SimpleArt Yes, I am definitely interested in any alternative (and smarter) way to evaluate this kind of series. $\endgroup$ – derpy Jan 12 '17 at 12:50
  • $\begingroup$ m.wolframalpha.com/input/… $\endgroup$ – Simply Beautiful Art Jan 12 '17 at 13:04

We may use the following

Lemma. $$ g_1(x)=\sum_{n\geq 0}\frac{\sin((2n+1)x)}{2n+1} $$ is the Fourier series of a rectangle wave that equals $\frac{\pi}{4}$ over $(0,\pi)$ and $-\frac{\pi}{4}$ over $(\pi,2\pi)$.

By integrating twice $g_1(x)$, we get that $$ g_3(x) = \sum_{n\geq 0}\frac{\sin((2n+1)x)}{(2n+1)^3}$$ is the Fourier series of a piecewise-polynomial function, that is odd and equals $\frac{\pi x(\pi -x)}{8}$ over the interval $(0,\pi)$. That proves your conjecture. I believe you my find interesting to also know that $$\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{E_{2m}}{2\cdot(2m)!}\left(\frac{\pi}{2}\right)^{2m+1}$$ that is proved here.

  • 1
    $\begingroup$ Thank you very much, this is pretty much as clean a solution as I could hope for. The fact that a closed formula can be written for each odd power is especially pleasing. $\endgroup$ – derpy Jan 12 '17 at 15:53

Using the Fourier series for the Bernoulli polynomials we have

$$\sum\limits_{k=1}^\infty \frac{\sin(2\pi x k)}{k^3}=\frac{(2\pi)^3 B_3(x)}{2\cdot 3!}$$

for $\enspace 0\leq x<1\enspace $ and with the Bernoulli polynomial $\enspace \displaystyle B_3(x)=x^3-\frac{3}{2}x^2+\frac{1}{2}x$ .

It follows (as assumed):

$\hspace{1cm}\displaystyle \sum\limits_{k=-\infty}^\infty \frac{\sin((2k+1)\pi x)}{(2k+1)^3}=2\sum\limits_{k=0}^\infty \frac{\sin((2k+1)\pi x)}{(2k+1)^3}= $

$\hspace{5.3cm}\displaystyle =2\left(\frac{(2\pi)^3 B_3(\frac{x}{2})}{2\cdot 3!}-\frac{1}{2^3}\frac{(2\pi)^3 B_3(x)}{2\cdot 3!}\right)=\frac{\pi^3}{4}x(1-x)$

Additional informations:

The general case of the Fourier series for the Bernoulli polynomials is $$\sum\limits_{k=1}^\infty \frac{\operatorname{Im}(i^{m-1}e^{i2\pi xk})}{k^m}=\frac{(-2\pi)^m B_m(x)}{2\cdot m!}$$
for $m\in\mathbb{N}$, $0\leq x <1$ and $x\ne 0$ for $m=1$ .

In the literature are often used the Euler numbers $E_{2m}$ instead of $-\frac{4^{2m+1}}{2m+1}B_{2m+1}(\frac{1}{4})$

which are defined by $\frac{1}{\cosh x}=\sum\limits_{k=0}^\infty\frac{x^k}{k!}E_k$ .

  • $\begingroup$ Thank you for your answer. I was not familiar with this class of polynomials. $\endgroup$ – derpy Jan 12 '17 at 15:56
  • $\begingroup$ @derpy : You are welcome. :-) $\endgroup$ – user90369 Jan 12 '17 at 19:59

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