# Evaluate $\int_0^1 \frac{1}{1+x^m}dx$ for $m>0$

I came across the sum: $$I(m)=\sum_{n=0}^\infty \frac{(-1)^n}{mn+1}$$ where $m>0$.

It's easy to see that this sum is equal to: $$\int_0^1 \frac{1}{1+x^m}dx$$ for $m>0$

So I tried my hardest to find a way of evaluating it. First I thought using Euler's Beta Function but to no avail. Then I decided to brute force a partial fraction decomposition, and I got this ugly mess (which is more than like incorrect):

$$I(m)=\sum_{n=0}^\infty \frac{(-1)^n}{mn+1} = \sum_{k=0}^\infty A_k\ln|1-e^{-(i(2k+1))/m}|$$Where $A_k$ equals

$$A_k= \frac{1}{\prod_{n=0}^k(e^{i(2k+1)}-e^{i\pi(2n+1)/m})\prod_{n=k}^\infty(e^{ik\pi/m}-e^{i(2n+3)}) }$$

I would check my work, but it's really tricky to check so I'm just hoping it's right. Is there a way of evaluating the integral that's less ugly than what I did?

• After looking at the $A_k$ I'm sure it's wrong. I wanted when k=0 for the first infinite product to be equal to 1. Oh well. – Tom Himler Jul 18 '18 at 3:06

Using the digamma function $$\int_0^1 \frac{1}{1+x^m}dx=\frac{\psi ^{(0)}\left(\frac{m+1}{2 m}\right)-\psi ^{(0)}\left(\frac{1}{2 m}\right)}{2 m}$$
Using the Hurwitz-Lerch transcendent function $$\int_0^1 \frac{1}{1+x^m}dx=\frac{\Phi \left(-1,1,\frac{1}{m}\right)}{m}$$