# improper integral: converges/diverges before evaluation?

• Mar 2nd 2013, 10:20 AM
infraRed
improper integral: converges/diverges before evaluation?
Hi. I'm a little confused about an issue in my homework problems involving improper integrals. I have problems like so:

Quote:

Determine whether the improper integral diverges or converges. Evaluate the integral if it converges.

$\displaystyle \int_{1}^{\infty}{\frac{ln\,x}{x}\,dx}$
How do I determine whether it converges or diverges before actually evaluating it?
• Mar 2nd 2013, 10:28 AM
Plato
Re: improper integral: converges/diverges before evaluation?
Quote:

Originally Posted by infraRed
Hi. I'm a little confused about an issue in my homework problems involving improper integrals. I have problems like so:
How do I determine whether it converges or diverges before actually evaluating it?

What is the derivative of $\displaystyle \frac{(\ln(x))^2}{2}~?$
• Mar 2nd 2013, 11:27 AM
infraRed
Re: improper integral: converges/diverges before evaluation?
ln(x)/x

...?
• Mar 2nd 2013, 11:34 AM
MINOANMAN
Re: improper integral: converges/diverges before evaluation?
Obviously diverges.
• Mar 2nd 2013, 11:51 AM
Plato
Re: improper integral: converges/diverges before evaluation?
Quote:

Originally Posted by infraRed
ln(x)/x...?

So what that tell you about the integral and why?
• Mar 4th 2013, 10:49 PM
SworD
Re: improper integral: converges/diverges before evaluation?
$\displaystyle \frac{\ln(x)}{x} > \frac{1}{x}$. But the integral of 1/x diverges, so the integral of the original function does too.