4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Tom Himler Jul 18 '18 at 3:06
4
$\begingroup$

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 $$\int_0^1 \frac{1}{1+x^m}dx=\frac{\Phi \left(-1,1,\frac{1}{m}\right)}{m}$$

$\endgroup$
  • $\begingroup$ That’s good enough for me, thank you. $\endgroup$ – Tom Himler Jul 18 '18 at 11:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.