# Integration by parts if f' ang g' are not continuous

1. May 30, 2012

### Boorglar

The Integration by Parts Theorem states that if f' and g' are continuous, then

∫f'(x)g(x)dx = f(x)g(x) - ∫f(x)g'(x)dx.

My question is, are those assumptions necessary? For example, this holds even if only one of the functions has a continuous derivative (say f' is not continuous but g' is) since in this case the right side can be differentiated using the FTC (since f*g' is continuous) and will yield f'*g, thus being an antiderivative of f'*g.

If BOTH f' and g' are discontinuous, is there an example for which this theorem does NOT work?
I tried finding one but I couldn't... Or is it still true, but much harder to prove?

A similar question arises with the substitution rule. They assume continuity of g' in the expression f(g(x))*g'(x)

2. May 30, 2012

### micromass

Staff Emeritus
The result also seems to be true (for indefinite integrals) in the more general case that $f^\prime$ and $g^\prime$ are merely Riemann-integrable. This restriction is obviously necessary, since if $f^\prime$ were not integrable, then there would be nothing to guarantee that $f^\prime g$ is integrable. So the integrals might not make sense.

Very nice and elegant theorems can be obtained if we extend the theory of integrals a bit. If we take Henstock integrals instead of Riemann integrals, then we can obtain the following theorems

If f and g are differentiable. Then f'g is Henstock-integrable if and only if fg' is. In that case
$$\int_a^b f^\prime g= fg\vert_a^b - \int_a^b fg^\prime$$

3. May 30, 2012

### Vargo

According to G. Folland's "Real Analysis", it is sufficient that f and g be absolutely continuous. In that case the derivative exists a.e. and the fundamental theorem of calculus holds.

You can weaken it slightly by instead assuming that f and g are of bounded variation. However, the result changes slightly depending in part on whether there are any points where both f and g are discontinuous.