Divergence/Convergence $\int_{5}^{\infty}e^{-\ln^{6}x}$ $$\int_{5}^{\infty}e^{-\ln^{6}x}$$
I have first tried to simplify the integrand 
$$e^{-\ln^{6}x}=\frac{1}{e^{\ln^{6}x}}=\frac{1}{e^{\ln x\cdot \ln^5x}}=\frac{1}{x^{\ln^5x}}$$
how should I continue? I am think about the limit test with some $\frac{1}{x^{\alpha}}$
 A: Since $\ln x$ is continuous on the domain considered it suffices to look at the integral from $e^2$ to $\infty$. But for $x\geq e^2$ we have $\ln x \geq 2$ so writing $(\ln x)^6 = (\ln x)^5 \ln x \geq 32 \ln x$: 
$$ \int_{e^2}^{\infty} e^{-(\ln x)^6} \; dx \leq \int_{e^2}^\infty e^{-32 \ln x} dx = \int_{e^2}^\infty \frac{1}{x^{32}} dx<+\infty$$
A: I suggest another path. By the change of variable, $u=\ln x$, $dx=e^udu$, one gets
$$
\int_{5}^{\infty}e^{-ln^{6}x}dx=\int_{\ln 5}^{\infty}e^{\large u-u^6}du
$$ then one may use that, as $u \to \infty$,
$$
u^2e^{\large u-u^6} \to 0
$$ to obtain, for some $u_0>\ln 5$,
$$
0<e^{\large u-u^6}<\frac1{u^2}, \quad u>u_0,
$$ yielding the convergence of the given integral.
A: If you wish to continue from where you left off, I can suggest the following: $$x\geq5\implies \ln x\geq\ln5\implies \ln^5x\geq\ln^5(5)>10$$
$$\frac{1}{x^{\ln^5(x)}}<\frac{1}{x^{10}}$$
$$\int_5^\infty \frac{1}{x^{\ln^5(x)}}dx< \int_5^\infty \frac{1}{x^{10}}dx=\left[\frac{x^{-9}}{-9}\right]_5^\infty=\frac1{17578125}$$
So the integral does converge...
A: Once we write the equivalent integral 
$$ I=\int_{\log 5}^{+\infty}\exp\left(u-u^6\right)\,du $$
from $\log 5>1$ we have that for every $u\geq \log 5$ the inequality $1-\frac{1}{u^5}\geq \frac{9}{10} $ holds, hence
$$ I \leq \int_{\log 5}^{+\infty}\exp\left(-\frac{9}{10}u^6\right)\,du\leq \int_{\log 5}^{+\infty}\frac{du}{1+\frac{9}{10}u^6}, $$
for instance.
