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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 ?

Yuriy S
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MathDisease
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    Use (or first prove) the fact that for $x$ sufficiently large, $\ln x<x^{1/3}$. – David Mitra Jun 03 '14 at 14:14
  • The primitive function is $-(2 \ln{x}+4) / \sqrt{x} + C$. From here the answer comes from Newton-Leibniz. The value of the integral is $\sqrt{2} (2 + \ln{2})$. – user153012 Jun 03 '14 at 14:15
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    Consider $$\int_2^\infty\frac{1}{x}dx.$$ In this expression, $x$ increases much faster than $1$, but the integral diverges. It's not about if something increases faster or not; sometimes intuitions are wrong. – Rocket Man Jun 03 '14 at 14:16
  • On a related note, $~\displaystyle\int_1^\infty\frac{\ln^ax}{x^{n+1}}dx=\frac{a!}{n^{a+1}}$ – Lucian Jun 03 '14 at 17:23

2 Answers2

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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.

DonAntonio
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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 $$\int udv=uv-\int vdu.$$

Rocket Man
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