TITLE: Is it possible to calculate $\int_{-1}^{1}\frac{1}{x}dx$ if the domain of $\frac{1}{x}$ is defined as "all reals except number zero"?
QUESTION [1 upvotes]: 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 
$$\int_{-1}^{1}\frac{1}{x}dx$$

REPLY [5 votes]: 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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