Convergent/Divergent Integral

**zer0e** · January 31st 2010, 10:39 PM

Evaluate the following integral if it is convergent.

$<br /> \int_1^{\infty}\frac{ln^2x}{x^3}dx$

**General** · January 31st 2010, 10:49 PM

Originally Posted by zer0e

Evaluate the following integral if it is convergent.

$<br /> \int_1^{\infty}\frac{ln^2x}{x^3}dx$

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

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

**Krizalid** · February 1st 2010, 06:21 AM

Of course we evaluate as long as having its convergence:

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

Thread: Convergent/Divergent Integral

LinkBack

Thread Tools

Display

Convergent/Divergent Integral

Similar Math Help Forum Discussions

Improper integral (convergent/divergent)

Improper integral convergent or divergent?

Improper integral. Convergent or divergent?

Help with convergent/divergent integral

Determing if the integral is convergent or divergent

Search Tags