Question: Integral : $$\int\_{2}^{\infty}\frac{\ln x}{x^{1.5}}dx$$

I noticed that this integral converges as $x$ goes to infinity since $x^{1.5}$ is much larger than $\ln x$ as $x$ increases, but I need to prove it in another way.Can someone show the proof ?

Answer: 
As hinted already in the comment, we have that $\;\forall\,\epsilon >0\;,\;\;\log x<x^\epsilon\;$ as long as we take $\;x\;$ big enough, and thus

$$\frac{\log x}{x^{1.5}}\le\frac{x^{0,1}}{x^{1.5}}=\frac1{x^{1.4}}\;\ldots$$

Of course, the above means you have to divide the integration interval in two, but one of them is a *finite*, nice regular Riemann integral, so it is only the improper one that matters to us.
Let $u=\ln x$ and let $dv=\frac{1}{x^{1.5}}dx$. Then $du=\frac{1}{x}dx$ and $v=\frac{(x^{-0.5})}{-0.5}$. Now use
$$\int udv=uv-\int vdu.$$