Show that F'(x) exists for all x [tex]\in[a,b] [/tex]

[irresistible]

Homework Statement

Consider
F(x) = x² sin(1/x²) if 0<x\leq1
and = 0 if x\leq0

Show that F'(x) exists for all x \in[a,b] but that F':[0,1] \rightarrow1 is not integrable.

Homework Equations

So we have to show we do not have F(1)-F(0) = \int F'(x)dx
(integral going from 0 to 1)

The Attempt at a Solution

I'm having trouble proving this statement.
Where should I start?
To show that F'(x) exists, should I just take the derivative or do I have to go under some long theorems of analysis to PROVE?
Thanks in advance.:shy:

 

[shaggymoods]

Most likely you'll have to actually *prove* that this derivative exists - it clearly exists for all x other than 0 because x^{2}sin(\frac{1}{x^{2}}) is the composition of differentiable functions. The only tricky part is whether F'(0) exists. To show that it does indeed exist you'll need to show that the following limit exists:

\lim_{h\to 0}\frac{f(h) - f(0)}{h}

But this just equals

\lim_{h\to 0}\frac{h^{2}sin(\frac{1}{h^{2}})}{h} = lim_{h\to 0}hsin(\frac{1}{h^{2}}) = 0

Then by application of the product rule and the chain rule,

F&#039;(x) = 2xsin(\frac{1}{x^{2}}) - \frac{2cos(\frac{1}{x^{2}})}{x^{2}}

if x is anything other than 0, and 0 if x=0. So all you need to show is that the above function is unbounded and it will consequently not be integrable.

 

[irresistible]

right,
But I think you made a little mistake with differentiation,
I think it will be:

F'(x)= 2x sin(1/x²) - 2cos(1/x²)/ x

Which still, limits exists.

But How do I prove that F'(x) is unbounded?

 

[HallsofIvy]

Take x= 1/\sqrt{2n\pi}. Then 1/x^2= 2n\pi so sin(1/x^2)= 0 and cos(1/x^2)= 1. F&#039;(x)= F&#039;(1/\sqrt{2n\pi})= -2\sqrt{2n\pi}. x will go to 0 as n goes to infinity. What happens to F'(x)?

 

[irresistible]

F'(x) will be Undefined as x approaches zero \rightarrow unbounded \rightarrow non integrable ?

 

[shaggymoods]

HallsofIvy just defined F'(x) above, so it *will* be defined; however, as n-->infinity, F'(x) also goes to infinity and this implies that the derivative is unbounded.

Also, thanks for pointing out my mistake; I was in an Econ class :P