Consider the sum (which is a Fourier series and a Dirichlet series):

$$F(x)=\sum_{n=1}^{\infty} n^{-s} \cos(2 \pi n x)$$

For $\Re(s)>1$ we have (thanks to absolute convergence) $\lim_{x \to 0} F(x)=\zeta(s)$ but does this holds for $0<\Re(s)<1$ ?

I am sure this is more than well known but I did not find it on internet, and as a simple Poisson summation formula does not answer the question, it is not completely easy, any reference ?

  • $\begingroup$ The difficulty here is that the zeta function isn't given by it's Dirichlet series representation over for real parts between $0$ and $1$. Rather its values are obtained via analytic continuation. $\endgroup$ – Joel Mar 4 '18 at 17:15
  • $\begingroup$ Yes, that is why I note $\zeta(s)$. $F(x)$ is well defined even for $0<\Re(s)<1$ when $x>0$ so what is the limit for $x \to 0$ in this case, do we still have this limit equal to $\zeta(s)$ by analytic continuation ? $\endgroup$ – Bertrand Mar 4 '18 at 17:40
  • $\begingroup$ The answer I posted was nonsense. Do you have some reason to think that $\lim_{x\to0}F(x)$ exists when $0<\Re s<1$? $\endgroup$ – David C. Ullrich Mar 4 '18 at 19:51
  • $\begingroup$ Analytic continuation may be ? If the limit does not exist then the queston is what is the asymptotic ? It is in litterature, for sure but I am not a professional and did not find it ! $\endgroup$ – Bertrand Mar 4 '18 at 21:12
  • $\begingroup$ @Bertrand After spending days on this problem, I found that the limit as $x\rightarrow 0+$ does not exist. But, this reference has much more information about polylogarithms, it contains a series expansion of $\sum \frac{e^{\mu}}{n^s}$ with $|\mu|<2\pi$ : cs.kent.ac.uk/pubs/1992/110 $\endgroup$ – Sungjin Kim Mar 15 '18 at 23:27

We shall see that the limit of $F(x)$ does not exist as $x\rightarrow 0^+$ for $0<s<1$. The following lemma follows from $$\cos(2\pi nx)\sin \pi x = \frac12 \left[ \sin \pi x(2n+1)-\sin \pi x(2n-1)\right]$$ and the telescoping sum.

Lemma 1

Let $t\ge 1$ and $x>0$. Then $$ A_t=\sum_{n\leq t} \cos (2\pi n x) =\frac{\sin\pi x(2\lfloor t\rfloor +1)-\sin\pi x}{2\sin \pi x}. $$

By partial summation, we have

Lemma 2

Let $x>0$ and $0<s<1$. Then $$ \sum_{n=1}^{\infty} \frac{\cos 2\pi nx}{n^s}=s\int_1^{\infty} \frac{A_t}{t^{s+1}}dt=-\frac12+s\int_1^{\infty} \frac{\sin\pi x(2\lfloor t\rfloor +1)}{2t^{s+1}\sin \pi x} dt. $$

We replace $\sin\pi x(2\lfloor t \rfloor +1)$ by $\sin 2\pi x t$ at a cost of a uniformly bounded function. By change of variable $x t = u$, we have

Lemma 3

Let $x>0$ and $0<s<1$. Then there are uniformly bounded functions $B_1(s,x)$ and $B_2(s,x)$ such that $$ \sum_{n=1}^{\infty} \frac{\cos 2\pi nx}{n^s}=B_1(s,x)+\frac s{1-s}B_2(s,x)+\frac{x^s}{2\sin \pi x} s\int_0^{\infty} \frac{\sin 2\pi u}{u^{s+1}}du.$$

By the integral in this post: I'm looking for several ways to prove that $\int_{0}^{\infty }\sin(x)x^mdx=\cos(\frac{\pi m}{2})\Gamma (m+1)$

we have an expression for the integral.

Lemma 4

For $0<s<1$, we have $$ s\int_0^{\infty} \frac{\sin 2\pi u}{u^{s+1}}du=(2\pi)^s \Gamma(1-s)\sin \frac{\pi s}2 $$

Back to the main problem, we have


For $x>0$ and $0<s<1$, and the uniformly bounded functions $B_i(s,x)$ the same as above lemma, we have $$ \sum_{n=1}^{\infty} \frac{\cos 2\pi nx}{n^s}=B_1(s,x)+\frac s{1-s}B_2(s,x) + \frac{x^s}{2\sin \pi x} (2\pi)^s \Gamma(1-s)\sin \frac{\pi s}2. $$

Therefore, for a fixed $0<s<1$, the limit of $F(x)$ as $x\rightarrow 0^+$ does not exist.


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