Analytical continuation of a Dirichlet series with periodic coefficients - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T18:52:59Z http://mathoverflow.net/feeds/question/26738 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26738/analytical-continuation-of-a-dirichlet-series-with-periodic-coefficients Analytical continuation of a Dirichlet series with periodic coefficients christian.selinger 2010-06-01T16:54:32Z 2010-06-01T20:11:34Z <p>Fix a complex number s and a real number x, does there exist an analytic continuation of the Dirichlet series </p> <p>$L(s,x):=\sum_{k=1}^{\infty}\frac{\sin^2(2\pi k x)}{k^s}$</p> <p>to the whole complex plane except 1?</p> <p>If yes, is there some functional equation verified which makes it possible to calculate $L(0,x)$?</p> <p>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.)</p> <p>Thanks for any comments Chri</p> http://mathoverflow.net/questions/26738/analytical-continuation-of-a-dirichlet-series-with-periodic-coefficients/26739#26739 Answer by Robin Chapman for Analytical continuation of a Dirichlet series with periodic coefficients Robin Chapman 2010-06-01T16:59:13Z 2010-06-01T19:10:33Z <p>Any Dirichlet series with periodic coefficients is analytically continuable to the whole plane (maybe with a pole at $1$).</p> <p>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 <em>Hurwitz zeta function</em> well-known to have an analytic continuation. <a href="http://en.wikipedia.org/wiki/Hurwitz_zeta_function" rel="nofollow">http://en.wikipedia.org/wiki/Hurwitz_zeta_function</a></p> <p><strong>Added</strong> 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.</p>