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I have to prove if this improper integral exist or not:

\begin{equation*} \int_{0}^{\infty} \frac{1}{x\sqrt{1+x}}dx \end{equation*}

And I found this theorem (here in the page four):

Let $\lim_{x\rightarrow \infty} x^{p}f(x)=c$, with c is a constant. Then:

  • $\int_{0}^{\infty}f(x)$ converges if $p>1$ and $c$ is finite;
  • $\int_{0}^{\infty}f(x)$ diverges if $p\leq1$ and $c\neq0$

So to solve it I proposed $p=\frac{3}{2}$, so we have $f(x)=\frac{1}{x\sqrt{1+x}}=\frac{1}{x^{3/2}\sqrt{\frac{1}{x}+1}}$

And, \begin{align*} \lim_{x\rightarrow \infty} \frac{x^{p}}{x^{3/2}\sqrt{\frac{1}{x}+1}}&=\lim_{x\rightarrow \infty} \frac{x^{3/2}}{x^{3/2}\sqrt{\frac{1}{x}+1}}\\ &=\lim_{x\rightarrow \infty} \frac{1}{\sqrt{\frac{1}{x}+1}}\\ &= 1 \end{align*}

So, as $p>1$, and $c$ is finite $\Rightarrow$ $\int_{0}^{\infty} \frac{1}{x\sqrt{1+x}}dx$ converges.

But! I saw in this question, that this improper integral doesn't converges. So, what am I doing wrong? Or how can I verify that it doesn't exist? I would really appreciate your help

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  • $\begingroup$ $\infty$ is not the problem $0$ is. $\endgroup$ – kingW3 Oct 6 '20 at 21:39
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    $\begingroup$ ${1 \over x}$ as $x \to 0^{+}$. $\endgroup$ – Felix Marin Oct 6 '20 at 21:40
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As noted on page $3$, there is one more assumption, which fails for your $f$. As it's not continuous at $x=0$, it's not continuous on $[0, \, \epsilon]$ for $\epsilon>0$.

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We have that

$$\int_{0}^{\infty} \frac{1}{x\sqrt{1+x}}dx=\int_{0}^{1} +\frac{1}{x\sqrt{1+x}}dx+\int_{1}^{\infty} \frac{1}{x\sqrt{1+x}}dx$$

and the first integral diverges by limit comparison test with $\int_{0}^{1} \frac{1}{x}dx$.

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