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let $x \in \mathbb{R} $,

how to determine the domain of convergence $\int_{1}^{+\infty}\frac{dt}{1+t^x}$ ,
my attempts :
I known if $x<0$ we get $\frac{1}{1+t^{x}}\rightarrow 1 [ t \rightarrow +\infty] $

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  • $\begingroup$ Break $\Bbb R$ into several intervals, then test convergence of the integral. $\endgroup$ – xbh Oct 1 '18 at 5:58
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    $\begingroup$ Could you fix your last formula? Also, note that even for $x=1$ the integral is divergent. My guess is that it's convergent on $(1,\infty)$. $\endgroup$ – Andrei Oct 1 '18 at 6:01
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    $\begingroup$ I think the OP means $$\frac 1{1+t^x} \to 1 [t \to +\infty].$$ $\endgroup$ – xbh Oct 1 '18 at 6:03
  • $\begingroup$ youtu.be/xro7c-mDk1g $\endgroup$ – Henry Lee Oct 1 '18 at 23:57
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If $0<x\leq 1$ then $\int_1^{\infty} \frac 1 {1+t^{x}} \, dt \geq \int_1^{\infty} \frac 1 {t^{x}+t^{x}} \, dt=\infty$ as seen by direct computation. For $x>1$ $\,$ $\int_1^{\infty} \frac 1 {1+t^{x}} \, dt \leq \int_1^{\infty} \frac 1 {t^{x}} \, dt <\infty$ again by direct computation.

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Also divergent for x=0 ; If x>0 and t>=1 then $\frac{1}{(t^x)+ (t^x)}$ <= $\frac{1}{1+t^x}$ <= $\frac{1}{t^x}$ so that the convergence domain is the same as for $\frac{1}{t^x}$ which you can integrate directly to see that the convergence domain is the interval (1,$\infty$)

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  • $\begingroup$ my problem is the demonstration because i can find the result $\endgroup$ – KEVIN DLL Oct 1 '18 at 6:21

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