# Improper Integral

• Dec 2nd 2008, 03:23 PM
sfgiants13
Improper Integral
Consider the integral: ln[x]/x^2 from 1 to infinity. Explain why it does not converge or evaluate the integral.

Using integration by parts, I eventually got [1/3]x^3 ln[x]-[1/9]x^3 from 1 to t as t approaches infinity. I know the answer is one, but using this I'll get infinity-infinity-0+1/9 when I substitute back in. Could someone help me out?
• Dec 2nd 2008, 04:02 PM
skeeter
$\displaystyle \int \frac{\ln{x}}{x^2} \, dx$

$\displaystyle u = \ln{x}$

$\displaystyle dv = \frac{1}{x^2} \, dx$

$\displaystyle du = \frac{1}{x} \, dx$

$\displaystyle v = -\frac{1}{x}$

$\displaystyle \int \frac{\ln{x}}{x^2} \, dx = -\frac{\ln{x}}{x} + \int \frac{1}{x^2} \, dx$

$\displaystyle \lim_{b \to \infty} \int_1^b \frac{\ln{x}}{x^2} \, dx =$

$\displaystyle \lim_{b \to \infty} \left[-\frac{\ln{x}}{x} - \frac{1}{x}\right]_1^b =$

$\displaystyle \lim_{b \to \infty} \left(-\frac{\ln{b}}{b} - \frac{1}{b}\right) - \left(-1\right) = 1$