Does $\int_{0}^{\infty} \frac{1}{x\sqrt{1+x}}dx$ converge? I have to prove whether this improper integral converges:
\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
 A: 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 $\epsilon>0$.
A: 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 $\int_{0}^{1} \frac{1}{x}dx$.
A: Well I think that the theorem isn't actually required to prove the divergence of the integral. We can simply find it's anti-derivative and show it's divergence.
$$\int_{}^{}\frac{1}{x\sqrt{1+x}}dx$$
$$x=\tan^{2}u \implies du=2\tan(u)\sec^{2}udu$$
$$2\int\frac{\tan(u)\sec^{2}(u)}{\tan^{2}(u)\sec(u)}du$$
$$2\int\sec(u)du$$
$$\ln(\frac{1-\cos(u)}{1+\cos(u)})$$
Note that $u=\arctan(\sqrt(x))$
$$\ln(\frac{1-\cos(\arctan(\sqrt(x)))}{1+\cos(\arctan(\sqrt(x)))})$$
Since the bounds of the integral mentioned in the question are from 0 to $\infty$, It's value at 0 is
$$\ln(\frac{1-\cos(\arctan(0)}{1+\cos(\arctan(0)})$$
$$\ln(\frac{1-1}{1+2})$$
$$\ln(0)=-\infty$$
So the integral in the required bounds diverges.
