Undergrad Integral resulting from the product of two functions/derivative functi

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The discussion centers on the possibility of defining two functions, f(x) and g(x), such that the integral ∫f(x)g'(x)dx lacks an indefinite solution while ∫f'(x)g(x)dx does possess one. Participants clarify that the term "not valid" refers to the integral not yielding results in terms of standard mathematical functions. A suggested example includes using g(x) = |x|, which can lead to complications in integration when 0 is within the range.

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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.
 
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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.
 
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Saracen Rue said:
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?
 
fresh_42 said:
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"
 
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.
 
fresh_42 said:
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)
 
I've encountered a few different definitions of "indefinite integral," denoted ##\int f(x) \, dx##. any particular antiderivative ##F:\mathbb{R} \to \mathbb{R}, F'(x) = f(x)## the set of all antiderivatives ##\{F:\mathbb{R} \to \mathbb{R}, F'(x) = f(x)\}## a "canonical" antiderivative any expression of the form ##\int_a^x f(x) \, dx##, where ##a## is in the domain of ##f## and ##f## is continuous Sometimes, it becomes a little unclear which definition an author really has in mind, though...

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