$\int_0^1 \frac{1}{1+x^k} dx$

Asked Sep 20 '24 at 10:54

Active Sep 20 '24 at 10:54

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I wanted to evaluate the following series: $$\sum_{k=0}^\infty (-1)^k\frac{1}{a+kb}$$

Where $a,b \in R$. After some manipulation(taking $1/a$ out and writing the general term of the sum as an integral from 0 to 1), I got that the series is equal to the following integral: $$\int_0^1 \frac{1}{1+x^k} dx$$

However I have no idea how to proceed from here on. I tried gamma and beta functions but they did not work , because the thing is $k \in R$, not $k \in Z$.

Is there any closed form of this integral as a function of k? And Also as a sidenote can we evaluate the sum given above in some other way?

asked Sep 20 '24 at 10:54

tensorman666

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In general, this is unknown. It can be phrased in terms of the Lerch phi function, of which some functional identities are known, but it's really just naming your series, not evaluating it. – Integrand Sep 20 '24 at 11:06
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The sum does not depend on $k,$ but the integral does. Seems like you missed something. – Thomas Andrews Sep 20 '24 at 11:37
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This is a sum that can be likely expressed with something such as the Hurwitz zeta or Dirichlet eta functions, but has no elementary closed form. – keixx Sep 20 '24 at 11:45
Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta – Anne Bauval Sep 20 '24 at 12:13
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This question is a megaduplicate: Values of $\int_{0}^{1}{\frac{dt}{1+t^n}}$, Evaluate $\int_0^1 \frac{1}{1+x^m}dx$ for $m>0 $, etc. – Anne Bauval Sep 20 '24 at 12:15
@AnneBauval Sorry, I did not know of this. – tensorman666 Sep 20 '24 at 14:52
Next time you can use Approach$0$: https://approach0.xyz/search/?q=AND%20site%3Amath.stackexchange.com%2C%20OR%20content%3A%24%5Cint_0%5E1%20%5Cfrac%7B1%7D%7B1%2Bx%5Ek%7D%20dx%24&p=1 – Anne Bauval Sep 20 '24 at 15:29

$\int_0^1 \frac{1}{1+x^k} dx$

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