1. Find covergence

Find if the improper integral converges. 1 / infinity lnx/x^2

2. Originally Posted by Link88
Find if the improper integral converges. 1 / infinity lnx/x^2
Consider that

$\frac{\ln(x)}{x^2}\leqslant\frac{\sqrt{x}}{x^2}=\f rac{1}{x^{\frac{3}{2}}}$

So what can we conclude?

3. do we have to use integration by parts? thats what my professor told us is the way to solve it.

4. Originally Posted by Link88
do we have to use integration by parts? thats what my professor told us is the way to solve it.
If you are trying to actually find the value of the integral then integration by parts is the correct method. To make it a little easier first try making the substitution $\ln(x)=z$

5. using subst. method i get u = lnx and then du = 1/x dx and substituting in i get u/x^2 = ln/x