Let G(x)=x^2 sin(π/x^2) for 0<x≤1 and G(0)=0. Show that the derivative g(x)=G’(x) exits but that g is not bounded so it is no integrable.
Let G(x)=x^2 sin(π/x^2) for 0<x≤1 and G(0)=0. Show that the derivative g(x)=G’(x) exits for all x in [0,1], but that g is not bounded so it is no integrable. (hence g has an antiderivative on [0,1] but is integrable on this interval). However, show that lim(a-->0+) ∫ (a to 1) g(x) dx exists.
At x=0, sin (pi/x^2) is not defined therefore,we can't use the differentiation rules to find the derivative at 0. Let's use the definition instead.
Lim_(x->0) |x|=0 therefore
Lim_(x->0)(G(x)-G(0))(x-0) exists and Lim_(x->0)(G(x)-G(0))(x-0)=0
For any `x!=0` G is the product and composition of differentiable function therefore G is differentiable and g(x) exists.
First answer: g(x) exists for any real x.
Let's find g(x) for x!=0.
`g(x)=2xsin(pi/x^2)- 2pix^2/x^3 cos(pi/x^2)`
Evaluate `g(x_n) ` for any `ninNN`
`Sin(pi/x_n^2)=0 and cos (pi/x_n^2)=1`
As `n->oo,` `x_n->0 and g(x_n)->-oo`
As `n->oo, y_n->0 and g(y_n)->+oo`
Therefore g has no lower bound nor upper bound in the neighborhood of 0.
In the neighborhood of 0, g has no lower bound nor upper bound therefore the upper rieman sum and lower riemann sum won't converge to the same value and g is not integrable on [0,1].
Between a>0 and 1 g is a well defined continuous function `int_a^1g(t)dt=G(1)-G(a)=0-a^2sin(pi/a^2)`
As `a->0, a^2->0`
Therefore the limit `lim_(a->0)` `int_a^1g(t)dt` exists and is 0.