Convergence of $\int_0 ^\infty \dfrac {dx}{\sqrt {1+x^3}}$
Attempt: $\lim_{x \rightarrow \infty} \dfrac {x^{\frac{3}{2}}}{\sqrt {1+x^3}} =1$
Hence, $\dfrac {1}{x^{\frac{3}{2}}}$ and $\dfrac {1}{\sqrt {1+x^3}}$ converge/diverge together.
$\int_0 ^\infty \dfrac {dx}{x^{\frac{3}{2}}}$ does not converge since, it's not defined at $0$.
However, my textbook says this integral is convergent. Why is it so?
Thank you for reading through!
On
For $x$ close to zero the initial function is bounded, hence it remains to verify convergence on $[T,+\infty)$ for large $T$. And $$ \int_T^\infty \frac1{x^{3/2}}\,dx $$ converges.
On
$1/(\sqrt {1+x^3})$ when x close to $0$ we can aproximate as 1+$\binom{{-1}/2}{1}x^{3}$ and this function is integrabile in interval [$0,a],\ a>0$. in $\infty$ you have aproximation integrabile function as $\dfrac {1}{x^{\frac{3}{2}}}$ is integrabile in [a,$\infty$].
On
Note that $f(x) = (1+x^3)^{-1/2}$ on $x \in [0,\infty)$ is bounded below by $g(x) = 0$, and above by $$h(x) = \begin{cases} 1, & 0 \le x \le 1, \\ x^{-3/2}, & x > 1. \end{cases}.$$ Therefore, $$0 < \int_{x=0}^\infty f(x) \, dx < \int_{x=0}^\infty h(x) \, dx = 1 + \int_{x=1}^\infty x^{-3/2} \, dx = 3.$$
An image for your convenience:
You got some good answers but I want to highlight some of the reasoning errors you made. It's important to derive your assertions carefully.
Remember that convergence is a property of the improper integral, not the function being integrated. So a statement like “$\frac{1}{x^{3/2}}$ converges” is ambiguous. What you mean is
This is true. Now for this:
A couple of problems here. One is that you have two improper integrals in one. Both $\int_0^1 \frac{dx}{x^{3/2}}$ and $\int_1^\infty \frac{dx}{x^{3/2}}$ are improper. But only the second one is relevant for this problem. That's because $\int_0^1 \frac{dx}{\sqrt{1+x^3}}$ is not improper, so you are only worried about whether $\int_1^\infty \frac{dx}{\sqrt{1+x^3}}$ converges.
As for $\int_0^1 \frac{dx}{x^{3/2}}$, you are correct that it does not converge, but it does not follow from your assertion that the integrand is not defined at $0$. After all, $\int_0^1 \frac{dx}{x^{1/2}}$ does converge, but the integrand is not defined at $0$ either. The divergence of $\int_0^1 \frac{dx}{x^{3/2}}$ follows from the $p$-test: if $p > 1$ then $\int_0^1 \frac{dx}{x^p}$ diverges and $\int_1^\infty \frac{dx}{x^p}$ converges.
To piece this all together: You know $\int_1^\infty \frac{dx}{x^{3/2}}$ converges by the $p$-test. Therefore, $\int_1^\infty\frac{dx}{\sqrt{1+x^3}}$ converges. Finally, $$ \int_0^\infty \frac{dx}{\sqrt{1+x^3}} = \int_0^1 \frac{dx}{\sqrt{1+x^3}} + \int_1^\infty\frac{dx}{\sqrt{1+x^3}} $$ so the given integral converges.