$\sum_{k=0}^\infty \frac{(-1)^k}{kn+1}$ and $\int_0^1 \frac{dx}{x^n+1}$

Question

Consider the infinite sum $\sum_{k=0}^\infty \frac{(-1)^k}{kn+1}$. I tried to solve it (like the Leibniz series) as an integral $$\int_0^1 \frac{dx}{x^n+1}$$ We have $$\frac{1}{x^n+1} = \prod_{k=0}^{n-1} \frac{1}{x+\omega^k\zeta} = \sum_{k=0}^{n-1} \frac{-\omega^k\zeta}{x+\omega^k\zeta}$$ with $\omega=e^\frac{2i\pi}n$ and $\zeta=e^\frac{i\pi}n$. We can then switch integration and summation to get -$$\frac1n \sum_{k=0}^{n-1} \int_0^1 \frac{\omega^k\zeta}{x-\omega^k\zeta}dx = -\frac1n \sum_{k=0}^{n-1} \omega^k\zeta\left|\ln y\right|_{y=-1}^{\frac1{\omega^k\zeta}-1}$$ after substitution $x=\omega^k\zeta(y+1)$. Expanding the difference, we get one term $\ln(-1)=i\pi$, which gets cleared because $\sum_{k=0}^{n-1} \omega^k\zeta=0$. This leaves $$-\frac1n\sum_{k=0}^{n-1} \omega^k\zeta\ln\left(\frac1{\omega^k\zeta}-1\right)=\frac{2i\pi\zeta}{n^2}\sum_{k=0}^{n-1} k\omega^k-\frac\zeta n\sum_{k=0}^{n-1}\left(\omega^k\ln\left(1-\omega^k\zeta\right)\right)$$

My two questions are now:
1. Is there a way to evaluate one of these sums into a closed form?
2. Is there another/an easier way to evaluate the title question?

Answer 1

$$\int_0^1 \frac{1}{x^n+1}\ dx = \sum_{k = 0}^{+\infty}(-1)^k \int_0^1 x^{nk}\ dx = \sum_{k = 0}^{+\infty} \frac{(-1)^k}{k n+1}$$

This is a well known function for those who manipulate Special function often, and in particular it's the Hurwitz-Lerch Phi function:

$$\sum_{k = 0}^{+\infty} \frac{(-1)^k}{k n+1} = \frac{\Phi \left(-1,1,\frac{1}{n}\right)}{n}$$

By basic knowledge in the field, this is a special case, which can be written as to Polygamma:

$$\frac{\Phi \left(-1,1,\frac{1}{n}\right)}{n} \equiv \frac{\psi ^{(0)}\left(\frac{n+1}{2 n}\right)-\psi ^{(0)}\left(\frac{1}{2 n}\right)}{2 n}$$

About the last sum

Notice that your sum is nothing but a general geometric series

$$\sum_{k = 0}^N k a^k = \frac{a \left(N a^{N+1}-(N+1) a^N+1\right)}{(a-1)^2}$$

You can obtain this by simply noticing that

$$k a^k = a\frac{d}{da} a^k$$

Answer 2

Due to $$ \sum_{n\geq 0}\frac{1}{(n+a)(n+b)}=\frac{\psi(a)-\psi(b)}{a-b}$$ we have $$ f(n)=\int_{0}^{1}\frac{dx}{1+x^n}=\sum_{k\geq 0}\frac{(-1)^k}{kn+1}=\frac{1}{2n}\left[\psi\left(\frac{1}{2n}+\frac{1}{2}\right)-\psi\left(\frac{1}{2n}\right)\right] $$ where the RHS can be expanded through the Gauss Digamma theorem for any $n\in\mathbb{N}$: $$ f(1)=\log(2),\quad f(2)=\frac{\pi}{4},\quad f(3)=\frac{\pi+\sqrt{3}\log(2)}{3\sqrt{3}},\quad f(4)=\frac{\pi+2\log(1+\sqrt{2})}{4\sqrt{2}}$$ and so on.