# domain of convergence $\int_{1}^{+\infty}\frac{dt}{1+t^x}$

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 $$\frac{1}{1+t^{x}}\rightarrow 1 [ t \rightarrow +\infty]$$

• Break $\Bbb R$ into several intervals, then test convergence of the integral. – xbh Oct 1 '18 at 5:58
• 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 $(1,\infty)$. – Andrei Oct 1 '18 at 6:01
• I think the OP means $$\frac 1{1+t^x} \to 1 [t \to +\infty].$$ – xbh Oct 1 '18 at 6:03
• youtu.be/xro7c-mDk1g – Henry Lee Oct 1 '18 at 23:57

If $$0 then $$\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$$ again by direct computation.
Also divergent for x=0 ; If x>0 and t>=1 then $$\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$$)