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....

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.

Asked on by rogersp44

1 Answer | Add Yours

Top Answer

cosinusix's profile pic

cosinusix | College Teacher | (Level 3) Assistant Educator

Posted on

Find g(0)

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.

 

(G(x)-G(0))(x-0)=x^2sin(pi/x^2)/x=xsin(pi/x^2)

0<=|(G(x)-G(0))(x-0)|<|x|

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

g(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)`

`g(x)=2xsin(pi/x^2)-2picos(pi/x^2)/x`

Let `x_n=1/sqrt(2n)`

Evaluate `g(x_n) ` for any `ninNN`

`pi/x_n^2=pi*2n`

`Sin(pi/x_n^2)=0 and cos (pi/x_n^2)=1`

`g(x_n)=-2pisqrt(2n)`

As `n->oo,` `x_n->0 and g(x_n)->-oo`

Let `y_n=1/sqrt(1+2n)`

`cos(pi/y_n^2)=-1, sin(pi/y_n^2)=0`

therefore `g(y_n)=2pisqrt(1+2n)`

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)`

`|int_a^1g(t)dt|<=a^2`

As `a->0, a^2->0`

Therefore the limit `lim_(a->0)` `int_a^1g(t)dt` exists and is 0.

We’ve answered 318,960 questions. We can answer yours, too.

Ask a question