Convergence of the integral $\int_0^\infty e^{-x^3} \mathop{\mathrm{d}x}$ I want to test the convergence of the integral $\int_0^\infty e^{-x^3} \mathop{\mathrm{d}x}$.
There are some parts of the solution which does not make sense to me, I'm hoping that someone can explain it to me.
The solution says:

The integrals $\int_0^\infty e^{-x^3} \mathop{\mathrm{d}x}$ and $\int_1^\infty e^{-x^3} \mathop{\mathrm{d}x}$ diverge and converge simultaneously. Therefore, $$\int_1^\infty e^{-x^3} \mathop{\mathrm{d}x} \leq \int_1^\infty e^{-x} \mathop{\mathrm{d}x} = 1.$$
  We conclude that $\int_0^\infty e^{-x^3} \mathop{\mathrm{d}x}$ converges by the comparison principle of Question $11$.

For reference, Question $11$ says:

Let $f$, $F$ be continuous on $(a,b)$. If the improper integral $\int^b_a F(x) \mathop{\mathrm{d}x}$ converges and $|f(x)| \leq F(x)$ for all $x:a < x < b$, then the improper integral $\int^b_a f(x)$ converges as well. Here $-\infty \leq a < b \leq \infty$.

I really feel that the solution has made an error somewhere.
$1.$ What does it mean when it says that the integrals diverge and converge simultaneously?
$2.$ How does the inequality say anything about the convergence of $\int_0^\infty e^{-x^3} \mathop{\mathrm{d}x}$?
 A: *

*It means either they both converge or they both diverge. The reason is that $\int_0^\infty e^{-x^3} dx = \int_0^1 e^{-x^3} dx + \int_1^\infty e^{-x^3} dx$ and the first term on the right is a finite number.

*They are just using the comparison principle that was mentioned: $|e^{-x^3}| \leq e^{-x}$ on $[1,\infty)$ and $e^{-x}$ is integrable there so $e^{-x^3}$ is also integrable there.

A: I think we should read the first line as $\int_{0}^{+\infty}e^{-x^3}\,dx$ and $\int_{1}^{+\infty}e^{-x^3}\,dx$ are both divergent or convergent. Anyway, through the substitution $x^3=z$ it follows that:
$$ \int_{0}^{+\infty} e^{-x^3}\,dx = \frac{1}{3}\int_{0}^{+\infty} z^{-2/3} e^{-z}\,dz = \frac{1}{3}\,\Gamma\left(\frac{1}{3}\right) = \color{red}{\Gamma\left(\frac{4}{3}\right)} $$
that is a number less than one but close to one, since $\Gamma(1)=\Gamma(2)=1$ and the $\Gamma$ function is a convex function with a bounded derivative on the interval $[1,2]$. A quite good approximation of $-\log\Gamma(x)$ over such interval is given by $\frac{(2-x)(x-1)}{2}$, hence:
$$ \int_{0}^{+\infty} e^{-x^3}\,dx \approx \exp\left(-\frac{1}{9}\right)\approx\frac{8}{9}.$$
A: *

*I think the sentence "diverge and converge simultaneously" means that either they both converge or they both diverge.

*Plug in $f(x)=e^{-x^3}$ and $F(x)=e^{-x}$, with bounds $a=1, b=\infty$
A: (1) $$\int_0^\infty e^{-x^3}\,dx = \int_0^1 e^{-x^3}\,dx  + \int_1^\infty e^{-x^3}\,dx $$
(2) As the integrand is positive, $R\mapsto\int_1^R e^{-x^3}\,dx$ is increasing, so the $\lim_{R\to\infty}$ exists and is finite or is infinite, but the second case is impossible (why?).
