Convergent/Divergent Integral

• Jan 31st 2010, 11:39 PM
zer0e
Convergent/Divergent Integral
Evaluate the following integral if it is convergent.

$
\int_1^{\infty}\frac{ln^2x}{x^3}dx$
• Jan 31st 2010, 11:49 PM
General
Quote:

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

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

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

To solve the integral, use integration by parts with $\color{blue}u=ln^2(x)$ and $\color{blue}dv=\frac{1}{x^3}dx$
• Feb 1st 2010, 07:21 AM
Krizalid
Of course we evaluate as long as having its convergence:

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