A complex exp-series and related trig-integral

[DreamWeaver]

This is not a tutorial... Just for a bit of fun. All contributions welcome! (Handshake)Here's an interesting series to explore. Let $$j\in\mathbb{N}\cup \{0\}$$, and $$p\in\mathbb{N}\ge 2$$, then consider the following series:$$\sum_{k=0}^{\infty}\frac{e^{2\pi ik/p}}{(2k+1)^{j+1}}$$As I say, it's just for fun. I'll post some results soon, but by all means, do help yourselves in the mean-time. Oh, go on! You know you want too... (Tmi)Relevance:

Spoiler

This complex-exponential series is essential for evaluating the following trigonometric integral, with $$m\in\mathbb{N}$$:$$\int_0^{\pi/p}\frac{x^m}{\sin x}\, dx$$

Series hint:

Spoiler

Split the series into real and imaginary parts, then express as Polygamma/Hurwitz Zeta functions.

 

[alyafey22]

Consider

$$\text{Li}_j(e^{\frac{\pi i}{p}})=\sum_{k\geq 1}\frac{e^{\frac{k\pi i}{p}}}{k^j}$$

By splitting into even and odd we have

$$e^{\frac{\pi i}{p}}\sum_{k\geq 0}\frac{e^{\frac{2k\pi i}{p}}}{(2k+1)^j}=\text{Li}_j(e^{\frac{\pi i}{p}})-2^{-j}\text{Li}_j(e^{\frac{2\pi i}{p}})$$

This Can be further sumplified using

$$Li_j(x)+Li_j(-x)=2^{1-j}Li_j(x^2)$$

 

[alyafey22]

Hey DW , I hope you can verify my results. I don't have the resources to do so.

 

[DreamWeaver]

Hey DW , I hope you can verify my results. I don't have the resources to do so.

Looks good to me, Zaid! But I have had mucho beer tonight, so I'll take a proper look tomorrow, when I find my brain again. I'm sure it's here somewhere... (Bandit)