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I was wondering how one could find the values of $n$ such that the area between $0$ to infinity under the graph $\frac{1}{x^n+1}$ converges. So far I've worked out that it does converge when $n = 2$ because it just becomes tan inverse which converges. For $n = 1$ however, it will diverge, as it becomes a log when integrated. For anything less than $0$ it will obviously also diverge.

I'm interested in the behavior of this area between $n=1$ and $n=2$ and I want to know how one could prove that there are other points of divergence bigger than $n=1$ and how you could find them. Or otherwise, prove there are no other points of divergence.

Any help with this question is much appreciated.

Gary
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Tom
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    The integral can be split into the integral over $[0,1]$ and then the integral over $[1,\infty]$. The first part is always finite. The second part is finite when $n>1$ by comparison with the integral of $1/x^n$. It is infinite when $n<1$ by comparison with $1/(2x^n)$ – Calvin Khor Jun 15 '23 at 03:14
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    If $n>1$, the area is $\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)$. See https://math.stackexchange.com/q/4459448 – Gary Jun 15 '23 at 03:32
  • See also https://math.stackexchange.com/questions/48740/int-0-infty-fracdx1xn – Travis Willse Jun 15 '23 at 03:36

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