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I was reading something which I do not understand.

I saw a conclusion that says that for any function that has a removable discontinuity or jump discontinuity (step discontinuity) then the function has no Indefinite integral.

So I took a look at $ \frac {1}{n} $

So as I saw the function suffer from jump discontinuity in $x = 0$.

But still as you can see it has Indefinite integral which is $\log(n)$.

So was something wrong in the statement I read or I misinterpreted the $\frac {1}{n}$ or the statement

Thanks in advanced !!

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The discontinuity at $x=0$ is not a jump discontinuity. A jump discontinuity must have both its left hand limit and its right hand limit be finite. –  Ataraxia Aug 14 '13 at 8:56
    
$\log x$ is not an antiderivative of $1/x$ on any neighbourhood of $x=0$. –  mrf Aug 14 '13 at 9:13
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2 Answers

You need to understand the theorem that derivatives cannot have jump discontinuities. This is a consequence of the Mean Value Theorem and has been discussed elsewhere on math.stackexchange.com

About your example the function $1/x$ does not have a jump discontinuity, but rather an infinity at $ x = 0$ so it does not apply here.

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Just because there is a jump discontinuity does not mean the function does not have an indefinite integral. Just take the integral of the respective functions with exclusion of the value that it's jumped at.

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The concept of an indefinite integral (anti-derivative) of $f(x)$ implicitly assumes that there is some interval in which we have a function $g(x)$ such that $g'(x) = f(x)$ on that interval. So strictly speaking when we talk about indefinite integral we should also take note of the interval in which it is valid. –  Paramanand Singh Aug 14 '13 at 9:07
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