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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

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1 Answer 1

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.

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.

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Alternatively, $f_y(s)$ can be expressed as a polylogarithm. –  Fredrik Johansson Jun 1 '10 at 20:36

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