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The following true or false question is taken by an Integral Calculus Exam but I cannot come up with a counter example.

Question: If a function has only one point of discontinuity , then it is Riemann integrable and its indefinite integral is a differentiable function.

Now, since our function has only point of discontinuity this means that it is Riemann integral. Let us call $F$ the antiderivative and let $x_0$ be the point of discontinuity of our function. Clearly $F$ cannot be differentiable at $x_0$ since $f$ is not continuous there. Do we know anything else about $F$? I think the statement is false but I cannot come up with a counter example. Any help?

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The definition of the Riemann integral insists that the function be bounded. So $f(x) = 1/x$ shows the statement is false. We can't integrate over any interval containing $x=0$.

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  • $\begingroup$ OK.. so we take $f$ to be unbounded like $f(x)=\frac{1}{x}$ and the interval how exactly? E.g would the integral $(-2, 2)$ suffice? $\endgroup$ – Tolaso Sep 17 '16 at 15:56
  • $\begingroup$ Yes. The function is not Riemann integrable over that interval. $\endgroup$ – B. Goddard Sep 17 '16 at 15:57
  • $\begingroup$ Hmm.. something bother me here. The function $f(x)=1/x$ cannot be said continuous at $x=0$ since it is not defined there at all. However in that interval it is not Riemann integrable. But in our assumptions we have that our function has only a point of discontinuity. Our function $f(x)=1/x$ has none. $\endgroup$ – Tolaso Sep 17 '16 at 16:00
  • $\begingroup$ No, $x=0$ is a point of discontinuity of $1/x$. $\endgroup$ – B. Goddard Sep 17 '16 at 16:01
  • $\begingroup$ Why? Our function function is not defined there at all. Throughout its domain which is $\mathbb{R}^*$ well it is continuous. $\endgroup$ – Tolaso Sep 17 '16 at 16:02

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