Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Fix a complex number s and a real number x, does there exist an analytic continuation of the Dirichlet series

$L(s,x):=\sum_{k=1}^{\infty}\frac{\sin^2(2\pi k x)}{k^s}$

to the whole complex plane except 1?

If yes, is there some functional equation verified which makes it possible to calculate $L(0,x)$?

If yes, what about the modulus of continuity of $x\mapsto L(0,x)$? ($L(\frac{3}{2},x)$ seems to be a nice case.)

Thanks for any comments Chri

share|improve this question
add comment

1 Answer

Any Dirichlet series with periodic coefficients is analytically continuable to the whole plane (maybe with a pole at $1$).

It's a finite linear combination of series like $$\sum_{m=1}^\infty\frac1{(km+r)^s}$$ where $1\le r\le k$ which equals $$k^{-s}\sum_{m=1}^\infty\frac1{(m+r/k)^s}.$$ This latter sum is an example of a Hurwitz zeta function well-known to have an analytic continuation. http://en.wikipedia.org/wiki/Hurwitz_zeta_function

Added Looking carefully at your question, I note that despite your title, your series does not actually have periodic coefficients unless $x$ is rational. In general your $k$-th coefficient is $$a_k=\sin^2 2\pi kx=\frac{2-\exp(4\pi i x)-\exp(-4\pi i x)}4.$$ Thus your series can be expressed in terms of the Riemann zeta function and functions of the form $$f_y(s)=\sum_{n=1}^\infty\frac{\exp(2\pi iky)}{n^s}.$$ In effect this sort of function is dual to the Hurwitz zeta function, and it has an analytic continuation to the complex plane with a pole at $1$ proved in the same manner as the Hurwitz zeta function. In the Wikipedia page it has a brief appearance as essentially $\beta(x;s)$. One can express $f_y(1-s)$ in terms of the Hurwitz zeta function.

share|improve this answer
    
Alternatively, $f_y(s)$ can be expressed as a polylogarithm. –  Fredrik Johansson Jun 1 '10 at 20:36
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.