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.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?
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.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.
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.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?
An improper integral is an integral where either the upper or lower limit of integration is infinite, or where the integrand function is not defined at one or more points within the bounds of integration.
Unlike a standard definite integral, an improper integral cannot be evaluated using the fundamental theorem of calculus. Instead, it requires special techniques such as limit evaluation or transformation into a proper integral.
The bounds of an improper integral are important because they determine the behavior of the integrand function at the endpoints. In the case of -1 to 1, it means that the function may not be defined or may approach infinity at one or both of these points.
An improper integral converges if its limit as the upper and lower bounds approach infinity is a finite value. It diverges if the limit approaches infinity or negative infinity, or if the function is not defined at one or both of the endpoints.
While some techniques may be applicable to both types of improper integrals, there are also specific methods for evaluating definite and indefinite improper integrals. For example, for definite integrals, we often use limits to evaluate the integral, while for indefinite integrals, we may need to use substitution or integration by parts.