1. ## Convergent/Divergent Integral

Evaluate the following integral if it is convergent.

$\displaystyle \int_1^{\infty}\frac{ln^2x}{x^3}dx$

2. Originally Posted by zer0e
Evaluate the following integral if it is convergent.

$\displaystyle \int_1^{\infty}\frac{ln^2x}{x^3}dx$
$\displaystyle \color{blue}\int_1^{\infty}\frac{ln^2x}{x^3}dx$
=$\displaystyle \color{blue}\lim_{t\to\infty} \int_1^{t}\frac{ln^2x}{x^3}dx$.

To solve the integral, use integration by parts with $\displaystyle \color{blue}u=ln^2(x)$ and $\displaystyle \color{blue}dv=\frac{1}{x^3}dx$

3. Of course we evaluate as long as having its convergence:

For all $\displaystyle x\ge1$ is $\displaystyle \ln x<\sqrt x,$ thus $\displaystyle \ln^2x<x$ hence $\displaystyle \frac{\ln^2x}{x^3}<\frac1{x^2},$ and the integral converges since $\displaystyle \int_{x\ge1}\frac{dx}{x^2}<\infty.$