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Is following reasong correct?

We can't find definite integral on discontinious part of function. Yet as far as I understand, when we talk about continuity of a function we mean continuity over the domain of said function. So if we change domain we can make discontinous function continious. Like by excluding zero from domain of 1/x we will make said function continious over its domain. After this it would be possible to find $$\int_{-1}^{1}\frac{1}{x}dx$$

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  • $\begingroup$ A precondition for definition of Riemann integral $\int_{a} ^{b} f(x) \, dx$ is that $f$ is defined over $[a, b] $ and that $f$ is bounded on this interval. The problem here is that however you try to define $f(0)$ the function remains unbounded. $\endgroup$ – Paramanand Singh Oct 12 '18 at 9:57
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No.

It is not enough for a function to be continuous on a (non-closed) interval for it to be integrable over that same interval. The function must also be bounded. Your function is not bounded. The Riemann integral of that function does not exist, including or excluding $0$.

Another way of thinking is that a single point has no effect on an integral. That is, if $f$ is equal to $g$ on all points except one, the integral of $f$ and $g$ will be equal. This statement can actually be expanded to saying that if $f$ and $g$ are equal on all points except for a countable many of them, their integrals will be equal. If you know a bit of measure theory, then an even stronger statement applies, since the integrals will be equal if $f$ and $g$ differ only on a set with measure $0$.

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  • $\begingroup$ "The function must also be bounded." Doesn't it contradict to fact that we can find definite integral $\int_{-2}^{-1}\frac{1}{x}dx$? If you were right we wouldn't be able to calculate it because 1/x isn't bounded function. $\endgroup$ – user161005 Oct 11 '18 at 8:48
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    $\begingroup$ @user161005 The function $\frac1x$ is bounded on $[-2,-1]$. $\endgroup$ – 5xum Oct 11 '18 at 8:57
  • $\begingroup$ I see, you meant "bounded on given interval". After remembering graph of 1/x I agree that the function isn't bounded on [-1;1] $\endgroup$ – user161005 Oct 11 '18 at 9:53
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    $\begingroup$ @user161005 No, because you cannot integrate a function that is not defined on a subinterval of $[a,b]$. $\endgroup$ – 5xum Oct 11 '18 at 11:31
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    $\begingroup$ @user161005 Editing your question to the point at which an already posted answer becomes incorrect is usually not advisable. $\endgroup$ – 5xum Oct 12 '18 at 8:58

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