Integral resulting from the product of two functions/derivative functi

[Saracen Rue]

TL;DR Summary

Define f(x) and g(x) so that the integral of f(x)*g'(x) is NOT valid but the integral of f'(x)*g(x) is.

Hey, sorry for the cluncky title. It was rathet difficult to summarise what I'm talking about here.

I want to know if it's possible to define $f(x)$ and $g(x)$ in such a way that $∫f(x)g'(x)dx$ has no indefinite solution while $∫f'(x)g(x)dx$ does have an indefinite solution.

Any help is greatly appreciated :)

Thank you all for your time.

 

[mitochan]

Hi
Partial integral rule we write
$$\int^b_a fg'dx=[fg]^b_a - \int^b_a f'g dx$$
When left side integral and
$$[fg]^b_a =f(b)g(b)-f(a)g(a)$$
are definite, the right side integral is also definite.
Thus your require above fg be indefinite but I am afraid not.

 

[fresh_42]

Summary: Define f(x) and g(x) so that the integral of f(x)*g'(x) is NOT valid but the integral of f'(x)*g(x) is.

Hey, sorry for the cluncky title. It was rathet difficult to summarise what I'm talking about here.

I want to know if it's possible to define $f(x)$ and $g(x)$ in such a way that $∫f(x)g'(x)dx$ has no indefinite solution while $∫f'(x)g(x)dx$ does have an indefinite solution.

Any help is greatly appreciated :)

Thank you all for your time.

What do you mean by "not valid"? Is the absolute value as $g(x)$ not valid, since $g'(0)$ doesn't exist?

 

[Saracen Rue]

What do you mean by "not valid"? Is the absolute value as $g(x)$ not valid, since $g'(0)$ doesn't exist?

I meant "not valid" as to mean "have no results in terms of standard mathematical functions"

 

[fresh_42]

Riemann or Lebesgue? If it is Riemann, then $g(x)=|x|$ and a suitable $f(x)$ should do if $0$ is in the integration range.

 

[Delta2]

Riemann or Lebesgue? If it is Riemann, then $g(x)=|x|$ and a suitable $f(x)$ should do if $0$ is in the integration range.

Though not explicitly stated, I believe the statement implies that the derivatives are well defined in the range of integration, just that the integrals have some sort of problem. (by "not valid" I think the OP means that the integral doesn't have a closed form)