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I have a question regarding existence of indefinite integral.

Consider a function $f:\mathbb{R}\rightarrow \mathbb{R}$. Using the definition, this function is integrable on $[a,b]$ (i.e. the definite integral on $[a,b]$ exists) if $f$ is bounded on $[a,b]$ and the lower integral (supremum over partitions of the lower Riemann sum) is equal to the upper integral (infimum over partitions of the upper Riemann sum).

My question is: when does the indefinite integral of $f$ is well defined? Do we need $f$ integrable on any $[a,b]$?

Star
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1 Answers1

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It probably suffices to say that $\int_0^x f(t) dt$ exists for every $x$, since that implies integrability over any interval $[a, b]$. For once you know that exists, you can define $$ F(x) = \int_0^x f(t)~dt $$ and $F$ is an indefinite integral for $f$, by the fundamental theorem.

Any other indefinite integral for $f$ has the same derivative (everywhere) as $F$, and hence, by the mean value theorem, differs from $F$ by a constant.

Perhaps your question means "Is there a function $f$ that's not integrable over some interval $[a, b]$, and yet still has an indefinite integral?"

If you'll confirm that this is what you were really wondering, I'll try to answer.

John Hughes
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  • Provided $[0,x]$ belongs to the domain of $f$. –  Dec 10 '15 at 13:25
  • The answer then is: a sufficient condition is $f$ integrable on any $[a,b]$. It is harder to say whether it is necessary. Right? – Star Dec 10 '15 at 13:30
  • Yves: the domain of $f$ was specified as the whole real line. STF: yes, that's a good summary of my answer. – John Hughes Dec 10 '15 at 15:02