1- Does the integrand and its antiderivative necessarily have the same domain?

I'm talking about the antiderivative incase of definite integration.

The answer should be no as the domain usually changes after differentiating, however this means that some functions aren't 'definitely' integrable just because both of the anti-derivative and the integrand don't agree on the same domain.

this can be solved by following a law which states that the limits of any definite integral(the interval of integration) should be a subset in the intersection of the domains of the integrand and its anti-derivative ( I'm not entirely sure of this).

an example which came into mind is differentiating ln(x) giving (X)^(-1) : (they have different domains), however this doesn't confirm my point as differentiating ln|x| also gives (X)^(-1) : (they have the same domain), when we integrate (X)^(-1) we take the antiderivitave which agrees with the integrand's domain. the same also works for functions integrated into arcsech(f(x)) (not only ln(x) ).

2- Do all functions have an antiderivative?

My answer would be yes.

the integrability of a function depends on whether you are definitely integrating on an interval which is a subset from the domain or not, some functions (which have graphs looking more like scatter plots) have a tedious/impossible antiderivitave, but since the function just represents a bunch of points it might not be integrable over any interval as its domain doesn't include any intervals.

3- does the integrability of a function depend on discontinuous/indifferentiable points?

My answer would be: no as long as these points aren't excluded in the domain, vertical asymptotes and consequent discontinuities are excluded from the domain (right?).

4- is it okay that some functions have multiple different antiderivitaves?

Although this is true (an example is antideriving sech(x) ), I can't quite give it sense as different antiderivitaves give out different values for the same input, which means that depending on the way you integrate sech(x) you will get different areas meanwhile the graph of sech(x) will show one certain area in a given interval.

-Any answer correctly explaining 'any' of these questions will be accepted. -'definetly integrable' means 'by using definite integration' Thanks for taking your time to read this.

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    $\begingroup$ Please format your text to make it more readable! $\endgroup$ – Aditya Dua Feb 13 at 21:05
  • $\begingroup$ Sorry, I think it's better now. $\endgroup$ – user597368 Feb 13 at 21:17

Partial answer : $(1)$ The answer is no. The function $$f(x)=\frac{1}{\sqrt{x}}$$ is not defined at $x=0$, but the antiderivate $$F(x)=2\sqrt{x}$$ is defined at $x=0$

$(2)$ The antiderivate cannot always be expressed by elementary functions. But for every function (ignoring pathological cases) , an antiderivate exists, at least in suitable intervals.

$(4)$ The antiderivate is unique upto a constant. If $F(x)$ is an antiderivate of $f(x)$, $F(x)+c$ with $c\in \mathbb R$ are all possible antiderivates. Antiderivates of the same function can look different in some cases (ignoring the constant-problem), but they can only differ by a constant. Note that a function can have different looking representations.


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