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I want to seek the real value $a$ such that $\int_0^{\infty} \frac{1}{1+x^a} dx$ converges by using comparisons test. Here is my method:

$$\int_0^{\infty} \frac{1}{1+x^a} dx = \int_0^{1} \frac{1}{1+x^a} dx + \int_1^{\infty} \frac{1}{1+x^a} dx$$

The integral $\int_0^1 \frac{1}{1+x^a} dx$ exists for all real number $a$ trivially. Next, since $\frac{1}{1+x^a} < \frac{1}{x^a}$ and $\int_1^{\infty} \frac{1}{x^a} dx$ converges when $a>1$, we conclude $\int_1^{\infty} \frac{1}{1+x^a} dx$ converges as well when $a>1$.

For $a\leq0$, since $$\lim_{x\to\infty}\frac{1}{1+x^a} \not=0 $$ it implies $\int_1^{\infty} \frac{1}{1+x^a} dx$ doesn’t converge. Also, $\int_1^{\infty}\frac{1}{1+x} dx$ doesn’t converge as well.

In the end, I guess that $\int_1^{\infty} \frac{1}{1+x^a} dx$ doesn’t converge when $0<a<1$, but I can’t find a suitable function $g(x)$ such that $g(x)\leq \frac{1}{1+x^a}$ for all $x\geq1$ (or maybe $x$ is large enough) and $\int_1^{\infty} g(x) dx$ doesn’t converge. Can anyone give some hints?

Jim Lin
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    For $x\in[1,\infty)$ and $0<a<1$, $\frac{1}{1+x^{a}}\geq\frac{1}{x^{a}+x^{a}}.$ – Danny Pak-Keung Chan Apr 15 '23 at 04:51
  • How amazing! I can’t thank you enough. – Jim Lin Apr 15 '23 at 05:01
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    You wrote $\lim_{x\to+\infty}\frac{1}{1+x^a} \not=0\Longrightarrow \int_1^{+\infty} \frac{1}{1+x^a} dx$ does not converge. For fear that you has some misunderstanding, I think I may say that $\lim_{x\to+\infty}f(x) \not=0\not\Longrightarrow \int_1^{+\infty} f(x)dx$ does not converge. For instance let $f:\mathbb{R}{\ge 0}\rightarrow\mathbb{R}$, $f(x)= \begin{cases} n&n\le x\le n+\frac{1}{n^3},\ n\in\mathbb{Z}{\ge 1}\ 0&\text{otherwise} \end{cases}$ – Asigan Apr 15 '23 at 07:59
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    As for how to select a $g(x)$ for $a<0$, we note that if $a<0$ then $\lim\limits_{x\rightarrow+\infty}\frac{x^a}{1+x^a}=1\implies$ $\forall\epsilon\in (0,1)$ $\ \exists M$, $ \frac{x^a}{1+x^a}>1-\epsilon$ when $x\ge M$, i.e. $\frac{1}{1+x^a}>(1-\epsilon)\frac{1}{x^a}$ when $x\ge M$, and $\int_M^{+\infty} (1-\epsilon)\frac{1}{x^a}dx$ does not converge since $\int_M^{+\infty}\frac{1}{x^a}dx$ does not. When formally writing a proof we select a "concrete" $\epsilon$, for instance user Danny Pak-Keung Chan selected $\epsilon=\frac{1}{2}$ – Asigan Apr 15 '23 at 08:12
  • https://math.stackexchange.com/questions/1836043/limits-at-infinity-of-a-function-with-convergent-improper-integral?noredirect=1&lq=1

    Here is the property that I use to explain why this integral doesn’t converge when $a<0$. Did I have some misunderstanding?

     – Jim Lin Apr 15 '23 at 10:26
  • Very nice. This inference is indeed correct. But I would suggest you should write $$\lim\limits_{x\rightarrow+\infty} \inf_{t\ge x}\frac{1}{1+t^a}\ne 0$$ in your proof, instead of $\lim\limits_{x\to+\infty}\frac{1}{1+x^a} \not=0$. (By the way, you can use @Asigan to let others know you sended a reply) – Asigan Apr 15 '23 at 15:25

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