In which cases and why can't we find any primitive?
if our IQ is too low
All rational functions have an elementary anti derivative.
>>7017342
find the values where x3+x2+1=0, go complex, shift the zeroes by i \epsilon , find a contour which vanishes and use the residue theorem to inegrate.
All functions have an antiderivative, it is simply not always computable in terms of elementary functions
>>7017354
>All functions
no
>>7017359
why?
>>7017365
Please take a math class and learn what an integral is.
>>7017365
Weierstrass functions
>>7017354
If the function is absolutely continuous (on a compact interval), then there is an antiderivative. In any other cases, you can't know it.
>>7017371
You are retarded. The Weierstrass function is continuous, hence has an antiderivative.
>>7017365
Function has to be continuous.
>>7017378
Are you fucking retarded? Absolute continuity is a much stronger condition than continuity.
>>7017342
Try to break it down in simple elements.
>>7017382
>Are you fucking retarded?
Yes, I got confused with another property. Sorry.
>>7017381
>>7017397
how about f(x)=1 of x=0, f(x)=0 for every other x
this can't be the derivative of any g(x), because then f should assume every value between 0 and 1
>>7017354
>>7017359
>>7017365
>>7017367
>>7017371
>>7017378
>>7017380
>>7017381
>>7017382
>>7017397
If g is a Lebesgue-integrable function on some interval [a,b] and we define f as:
f(x) = \int_a^x g(t)dt
then:
1. f is absolutely continuous
2. f is differentiable almost everywhere
6. it's derivative coincides almost everywhere with g(x)
So, if g is Lebesgue-integrable then, almost everywhere, f' = g (i.e it has an antiderivative).
>>7017401
Your function f is a weak derivative of every constant function.
>>7017406
so?
why does that imply if f is a weak derivative of some g than there is a h such that f is the strong derivative of h?
we were looking for functions that dont have antiderivative
>>7017405
>almost everywhere
