Exact relation between Newton's and Riemann's integral

Question

I understand that it follows from the first fundamental theorem of calculus that if a function is continuous, it has a primitive function and the Newton's and Riemann's integral have the same value.

What about non-continuous functions? Is it true that if both integrals exist, they are the same? It can obviously happen that function is Newton integrable but not Riemann integrable (e.g. unbounded functions). What about the other way? Can a function be Riemann integrable but not Newton integrable?

I looked up many questions on this site but found nothing.

Answer 1

If a function has more integrals, all of them are the same.

For example, the function signum on a limited interval can be integrated by Riemann but not by Newton. It's because signum doesn't have any primitive function.

Answer 2

The other example (complementing Hume's):

There are examples of everywhere differentiable function $f$ such that the Riemann integral $\int_0^1 f'(x)\;dx$ does not exist (although the Newton integral trivially exists as $f(1)-f(0)$). In this example, the function $f'$ is discontinuous.

These examples are one motivation for the gauge integral.

Answer 3

Possessing an anti-derivative does not necessarily imply that the function is Riemann integrable and further being Riemann integrable does not necessarily imply that the function possesses an anti-derivative.

However the Fundamental Theorem of Calculus states that if a bounded function $f$ possesses an anti-derivative $F$ on interval $[a, b] $ and further if $f$ is Riemann integrable then $$\int_{a} ^{b} f(x) \, dx=F(b) - F(a) $$ Thus there are some functions which are integrable both in sense of Newton and Riemann and then their integrals are equal.