Improper Integrals: Definite & Indefinite | Bounds -1 to 1

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if I wanted to take the definite integral of 1/x with respect to x, with the bounds -1 and 1, the integral would be improper.

What about the indefinite integral? We can find the indefinite integral of 1/x to be ln|x|. Can we find the indefinite integral of discontinuous functions?
 
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FS98 said:
if I wanted to take the definite integral of 1/x with respect to x, with the bounds -1 and 1, the integral would be improper.

What about the indefinite integral? We can find the indefinite integral of 1/x to be ln|x|. Can we find the indefinite integral of discontinuous functions?
The function f(x) = ln|x| is defined and continuous on two disjoint intervals: ##(-\infty, 0)## and ##(0, \infty)##. An indefinite integral is an antiderivative, a function, while a definite integral represents a number. Since ln|x| is differentiable on either of the two intervals listed above, and its derivative is 1/x, then it's valid to say that ##\int \frac {dx} x = \ln|x|## plus a constant.
 
Mark44 said:
The function f(x) = ln|x| is defined and continuous on two disjoint intervals: ##(-\infty, 0)## and ##(0, \infty)##. An indefinite integral is an antiderivative, a function, while a definite integral represents a number. Since ln|x| is differentiable on either of the two intervals listed above, and its derivative is 1/x, then it's valid to say that ##\int \frac {dx} x = \ln|x|## plus a constant.
I am nitpicking here, but it seems to me that we had a discussion around this sort of topic some time ago. The antiderivative would be a family of functions which are definable piecewise as ##\ln|x|## plus a constant for x < 0 and ##\ln|x|## plus a [possibly different] constant for x > 0.
 
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FS98 said:
if I wanted to take the definite integral of 1/x with respect to x, with the bounds -1 and 1, the integral would be improper.

What about the indefinite integral? We can find the indefinite integral of 1/x to be ln|x|. Can we find the indefinite integral of discontinuous functions?
Continuity is not necessary for integrability. So, yes, we certainly can find antiderivatives for functions containing discontinuities. Whether you can express these antiderivatives as compositions of elementary functions is another matter, however.

Riemann integrals are the work of the devil, I tell you.