Antiderivative and integrability

[Corvinus]

Hello everyone,

I was wondering : how do you proove that a function like sgn(x) is integrable but has no antiderivative? I mean there exists functions that are not continuous but have an antiderivative but why this one doesn't? And how do you proove then that it is integrable? The integral is defined as the variation of the antiderivative at two points, right? Is there someone to help me clarifying my ideas?

Thanks in advance,

 

[HallsofIvy]

Perhaps I am not understanding what you are saying. the "sgn" function, sgn(x)= -1 if x< 0, sgn(0)= 0, sgn(x)= 1 if x> 0, certainly is integrable. Its "anti- derivative" is |x|. And it is not clear to me what distinction you are making between being "integrable" and having an "anti-derivative".

 

[Corvinus]

By having an antiderivative, I mean having a function F(x) such that F'(x) = sgn(x). But how could it be F(x) = |x| since it's not differentiable at x = 0? I'm not sure about what I'm telling you but my book says that this function on [-1;1] has an definite integral = 0 but has no antiderivative. But then why is it integrable?

 

[pka]

By having an antiderivative, I mean having a function F(x) such that F'(x) = sgn(x). But how could it be F(x) = |x| since it's not differentiable at x = 0? I'm not sure about what I'm telling you but my book says that this function on [-1;1] has an definite integral = 0 but has no antiderivative. But then why is it integrable?

I can give you an answer to your question. But unless you are in an honors calculus and/or in some sort of analysis course, the explication will most likely leave you cold.

In general one should be really comfortable working with partitions of an interval $\displaystyle \left[ {a,b} \right]$ and refinements of partitions. If $\displaystyle \mathcal{P}$ is a partition of $\displaystyle \left[ {-a,a} \right]$ where $\displaystyle a>0$. Suppose that $\displaystyle 0\notin\left[ {a_j,a_{j+1}} \right]$, a sub-interval of $\displaystyle \mathcal{P}$. Then both endpoints have the same sign and so any representative sum over that cell of the partition is , $\displaystyle \pm|[a_{j+1}-a_j|$. All of those add up to $\displaystyle -(a_k+a)~ or~ (a-a_{k+1})$
Say that $\displaystyle 0\in\left[ {a_k,a_{k+1}} \right]$. Now $\displaystyle a_k\le 0$ and $\displaystyle a_{k+1}>0$. Thus then the sum over that sub-interval is $\displaystyle |a_{k+1} -a_j|$. Now refine the partition so that $\displaystyle |-a-a_k|=|a-a_{k+1}|$ then those sums add to zero all that is left is a number $\displaystyle f(x)[a_{k-1}-a_k|$ which is as small as we want. Thus the integral equals zero.

You can go back and review Riemann-Sums.