Convergence of: ${\int} _{3}^{+\infty}xe^{-(\log x)^\alpha} dx$ for $\alpha \in \mathbb{R}$ Study the convergence of the following integral for $\alpha \in \mathbb{R}$
$$ \int_{3}^{+\infty}\frac{x}{e^{(\log x)^\alpha}} dx$$
I have never used L'Hôpital's rule for the convergence of improper integrals, but giving that:
$$ \int_{3}^{+\infty}\frac{x}{e^{(\log x)^\alpha}}\,dx = \lim_{M \to +\infty} \int  _{3}^{M}\frac{x}{e^{(\log x)^\alpha}}\,dx$$
and that any hypothesis is respected, by this way I can say that it diverges for $\alpha \leqslant 1$ and converges for $\alpha > 1$.
Are the other ways without using L'Hôpital's rule?
 A: let $x=e^u$ then integral is 
$$\int_{\ln3}e^{u-u^\alpha+1}du$$
when $\alpha<1$ and from $e^x>x$ then trivially
$$\int_{\ln3}e^{u-u^\alpha+1}du>\int_{\ln3}(u-u^\alpha+1)du\to\infty$$
A: Make a substitution $x=e^u$, the integral becomes
$$\int^\infty_{\ln 3} \exp(u-u^a+1)du$$
As an alternative form of p-test,
$$\int^\infty_1 \exp(f(x))dx$$
converges only if $\lim_{x\to\infty}\frac{f(x)}{\ln x}<-1$.
We need $a>1$ for convergence, because when $a>1$, $\lim_{u\to\infty}\frac{u-u^a+1}{\ln u}=-\infty$.
In addition, when $a=1$, $\lim_{u\to\infty}\frac{u-u^a+1}{\ln u}=0\not< -1$.
When $a<1$, $\lim_{u\to\infty}\frac{u-u^a+1}{\ln u}=+\infty\not<-1$.
Why does the alternative p-test work?
I think it is better to give an intuitive explanation. Firstly, for $f(x)=k\ln x$, p-test implies convergence only for $k<-1$. 
If another $f(x)$ diverges to $-\infty$ quicker than $-\ln x$, then $e^{f(x)}$ drops to zero quicker, and by comparison test the integral will converge. In this case, As $f$ goes negative more quickly, $f(x)<-\ln x$ for large $x$, meaning that $$\lim_{x\to\infty}\frac{f(x)}{-\ln x}>1$$
Similarly, if another $f(x)$ goes negative slower than/at the same rate as $-\ln x$, by comparison test the integral diverges. In this case, $f(x)\ge-\ln x$ for large $x$, meaning that $$\lim_{x\to\infty}\frac{f(x)}{-\ln x}\le 1$$
It can be rigorously proved that $$\int^\infty_1\exp{f(x)}dx\text{ converges iff }\lim_{x\to\infty}\frac{f(x)}{-\ln x}>1$$ provided that $f$ is continuous in $[1,\infty)$.
