Proving $\int_0^\infty e^{-x^p} dx$ converges or not How can we prove that the following integral:
$$\int_0^{+\infty} e^{-x^p} dx$$
converges or not?
($p$ is any given number)

*

*Based on what I've done so far, I think we should separate the cases $p \ge 1$ and $0 < p < 1$ in order to solve the question successfully.

 A: No, you should separate into cases involving $p \leq 0$ and $p > 0$.
If $p \leq 0$, then ${-x^p} \to l$ as $x \to \infty$ for some finite $l$, so $e^{-x^p} \to e^{l}$ as $x \to \infty$. There's no way the integral can exist in this case : if the integrand coverges to a non-zero value, the integral can't converge.
If $p >0$, then we will show that for every $n$, the integral $\int_{0}^\infty e^{-\sqrt[n]{x}} dx$ exists. By monotonicity, the integral will exist for all $p > 0$, since for every $p$ there is $N$, $p > \frac 1N$, and then you can use the domination.
So make a change of variable : $t = \sqrt[n]x$, then $t^n = x$ so $nt^{n-1}dt = dx$.
From here , we get $n\int_{0}^\infty t^{n-1}e^{-t} dt$. Write this as $n \int_0^{\infty} e^{(n-1)\log t - t} dt$.
We know that $(n-1)\log t - \frac{t}{2} \to -\infty$ as $t \to \infty$. Therefore, we can find $T$ such that $t>T$ implies that $(n-1)\log t - \frac{t}{2} <0$. Now, we get $$
n \int_0^{\infty} e^{(n-1)\log t - t} dt = n \int_0^{T}e^{(n-1)\log t - t} dt + n \int_{T}^{\infty}e^{(n-1)\log t - t} dt
$$
The first integral is bounded, because $t^{n-1}e^{-t}$ is continuous on this interval, hence bounded, therefore $$
n \int_0^{T}e^{(n-1)\log t - t} dt \leq nT \max_{[0,T]} t^{n-1}e^{-t}
$$
For the other integral, we note that if $t>T$ then $$
e^{(n-1)\log t - t} = e^{(n-1)\log t - \frac t2} e^{-\frac t2} < e^{-\frac t2} 
$$
Therefore $$
n \int_{T}^{\infty}e^{(n-1)\log t - t} dt < n \int_T^{\infty} e^{-t/2}dt = 2ne^{-T/2}
$$
Finally, we obtain $$
\int_{0}^{\infty}e^{\sqrt[n]{x}} dx <n \left[\max_{[0,T]} t^{n-1}e^{-t}\right] + 2ne^{-T/2}
$$
which concludes the problem.
However, you can do better. Prove for yourself the following relation using integration by parts :
$$
\forall k \geq 1  \quad\int_{0}^\infty t^{k}e^{-t}dt = k\int_{0}^\infty t^{k-1}e^{-t}dt
$$
From here, conclude by induction that in fact, $\int_{0}^\infty e^{-\sqrt[n]{x}}dx = 1 \times 2 \times ... \times n = n! < \infty$.
A: Assuming $p>0$, let 
$$x^p=t \implies x=t^{\frac{1}{p}}\implies dx=\frac{1}{p}t^{\frac{1}{p}-1}$$
$$\int e^{-x^p} dx=\frac{1}{p}\int e^{-t}\,t^{\frac{1}{p}-1}\,dt=-\frac{1}{p}\,\Gamma \left(\frac{1}{p},t\right)$$
$$\int_0^{+\infty} e^{-x^p} dx=\frac{1}{p}\int_0^{+\infty} e^{-t}\,t^{\frac{1}{p}-1}\,dt=\frac{1}{p}\,\Gamma \left(\frac{1}{p},0\right)\quad > \quad \forall p > 0$$
A: Suppose $p>0$. The function $x\mapsto e^{-x^p}$ is positive continuous on $[0,\infty)$, so it suffices to look at  its magnitude at infinity. We have $$e^{-x^p}=o\left(\frac{1}{x^2+1}\right)$$
as $x\to \infty$.
Since $$\int_{0}^\infty \frac{1}{x^2+1} dx$$ converges, it follows that  $$\int_{0}^\infty e^{-x^p} dx$$
converges as well.
If $p<0$, then $$\int_{0}^\infty e^{-x^p} dx\geq \int_{N}^{2N} e^{-x^p} dx\geq  \int_{N}^{2N} e^{-N^p} dx\geq Ne^{-N^p}\to \infty,$$
as $N\to \infty$.
A: I assume that
$ p > 0$
and use the result,
often proved here,
that
$\lim_{x \to \infty} \dfrac{\ln(x)}{x^c}
\to 0
$
for any $c > 0$.
I want to show that
$e^{-x^p}
\lt \dfrac1{x^2}
$
for large enough $x$.
This will show that
the integral converges.
Taking logs,
this is
$-x^p \le -2\ln(x)
$
or
$x^p \ge 2\ln(x)
$
or
$\dfrac{\ln(x)}{x^p}
\le \dfrac12$
and this holds since
the left side
goes to zero.
