# Convergeance of Improper Integral

• Jan 26th 2013, 07:40 AM
renolovexoxo
Convergeance of Improper Integral
The problem statement is:
Let of be defined on (0,1] by f(x)=d/dx(x2sin(1/x)=2xsin(1/x^2)-(2/x)cos(1/x^2)

Show the improper Riemann Integral of f on (0,1] converges, but that the improper integral of |f| diverges on (0,1].

I went through and solved the integral as an integral from c to 1, then took the limit as c approaches 0 for this function. For both I have found a number that they limit goes to as it approaches 0. I'm having trouble showing that the limit as |f|-> 0 diverges. I went through the same process with |f| as I did with f. Is there maybe a trick to this other than straight solving it?
• Jan 26th 2013, 06:15 PM
chiro
Re: Convergeance of Improper Integral
Hey renolovexoxo.

Can you show us what you did?
• Jan 27th 2013, 07:09 AM
renolovexoxo
Re: Convergeance of Improper Integral
I evaluated
lim c->0+ of (x^2*sin(1/x)) and then did the same for |f| by using the absolute value of the same function.