# Does $\int_{0}^{\infty} \frac{1}{x\sqrt{1+x}}dx$ converges?

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

• $\infty$ is not the problem $0$ is. – kingW3 Oct 6 '20 at 21:39
• ${1 \over x}$ as $x \to 0^{+}$. – Felix Marin Oct 6 '20 at 21:40

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 $$\epsilon>0$$.
$$\int_{0}^{\infty} \frac{1}{x\sqrt{1+x}}dx=\int_{0}^{1} +\frac{1}{x\sqrt{1+x}}dx+\int_{1}^{\infty} \frac{1}{x\sqrt{1+x}}dx$$
and the first integral diverges by limit comparison test with $$\int_{0}^{1} \frac{1}{x}dx$$.