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For $0 < \theta < 2\pi$, define $$\kappa(x,\theta) = \frac{1}{\zeta(x)}\sum_{n=1}^\infty \frac{e^{ in\theta}}{n^x}$$ for $\Re(x) > 1$. It is easy to see that $$\kappa(x,\theta) = \frac{1}{\zeta(x)\Gamma(x)}\int_0^\infty \frac{t^{x-1}}{e^{t - i\theta}-1}dt $$

My question is what the analytic continuation of this function $\kappa(x,\theta)$ looks like for $x=0$ and negative integer $x$. We can assume $\theta$ is fixed.

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The Clausen function $\,\operatorname{Cl}_x(\theta)\,$ is often considered with $x$ a positive integer but allow analytic continuation for all complex values of $x$ (except $\,x=1$ and $\,\theta\equiv 0\pmod {2\pi}$).
Concerning non integer values of $x$ rewrite them as polylogarithms may be useful :

The polylogarithm function verifies indeed $\:\operatorname{Li}_x\left(e^{i\theta}\right)=\displaystyle \sum_{n=1}^\infty \frac{e^{ in\theta}}{n^x}\;$ so that $$\kappa(x,\theta) = \dfrac{\operatorname{Li}_x\left(e^{i\theta}\right)}{\zeta(x)}$$

and you may obtain values and representations for fixed $\theta$ using W. Alpha (here $\theta=1$) :

theta=1

Concerning analytic extensions as $x=0$ and $\,x$ a negative integer they derive directly from $$\operatorname{Li}_0(z)=\sum_{k=1}^\infty \frac {z^k}{k^0}=\frac z{1-z}$$ at each step compute $\;\operatorname{Li}_{-n}(z)=z\,\dfrac d{dz}\operatorname{Li}_{-n+1}(z)$ to get : \begin{align} \operatorname{Li}_{\;0}(z)&= \frac z{1-z}\\ \operatorname{Li}_{-1}(z) &= \frac z{(1-z)^2}\\ \operatorname{Li}_{-2}(z) &= \frac {z\,(1+z)}{(1-z)^3}\\ \operatorname{Li}_{-3}(z) &= \frac {z\,(1+4z+z^2)}{(1-z)^4}\\ \end{align} while $\,\operatorname{Li}_{\,1}(z)=-\log(1-z)\,$ will be fine for $z\neq 1$.

Hoping this helped even if very late (I saw this only recently sorry...).

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  • $\begingroup$ Thanks a lot. It is very helpful. $\endgroup$ – Willie Wu Dec 29 '15 at 1:49
  • $\begingroup$ Glad it helped @WillieWu! Note too that $\dfrac{\operatorname{Li}_x\left(e^{i\theta}\right)}{\zeta(x)}$ will be singular for $x=-2n$ and $\theta\not\equiv \pi\pmod{2 \pi}$ from $\zeta(-2n)=0$ (for $\theta\equiv \pi\pmod{2 \pi}$ the ratio admits the limit $\,2^{2n+1}-1$). Excellent continuation, $\endgroup$ – Raymond Manzoni Dec 29 '15 at 11:26
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You can perhaps use the following (if I am not wrong): We have: $$\frac{t^{x-1}}{\exp(t-i \theta)-1}-\frac{t^{x-1}}{\exp(t)(\exp(-i\theta)-1)}=-\frac{\exp(t)-1}{t}\frac{t^x}{\exp(t)(\exp(t-i\theta)-1)(\exp(-i\theta)-1)}$$

Hence $$\int_0^{+\infty}\frac{t^{x-1}}{\exp(t-i \theta)-1}=\frac{\Gamma(x)}{\exp(-i\theta)-1}-g(x)$$ and $g(x)$ is well defined (and analytic ) for ${\rm Re}(x)>-1$.

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  • $\begingroup$ Your steps help. Thanks. $\endgroup$ – Willie Wu Oct 16 '15 at 9:35

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