Answers to the question of the integral of $\frac {1} {x}$ are all based on an implicit assumption that the upper and lower limits of the integral are both positive real numbers.

@user599310, I am going to attempt some pseudo math to show it: $$ I^2 = \int e^-x^2 dx \times \int e^-x^2 dx = Area \times Area = Area^2$$ We can replace one x, with a dummy variable, …

Nov 29, 2013 · Wolfram Mathworld says that an indefinite integral is "also called an antiderivative". This MIT page says, "The more common name for the antiderivative is the …

Aug 9, 2017 · So, I can intuitively grasp that the derivative of the integral of a given function brings you back to that function. Is this also the case with the integral of the derivative?

Feb 4, 2018 · The integral of 0 is C, because the derivative of C is zero. Also, it makes sense logically if you recall the fact that the derivative of the function is the function's slope, because …

If by integral you mean the cumulative distribution function $\Phi (x)$ mentioned in the comments by the OP, then your assertion is incorrect.

I was reading on Wikipedia in this article about the n-dimensional and functional generalization of the Gaussian integral. In particular, I would like to understand how the following equations are

2 days ago · For questions about the properties of integrals. Use in conjunction with (indefinite-integral), (definite-integral), (improper-integrals) or another tag (s) that describe the type of …

Feb 17, 2025 · The noun phrase "improper integral" written as $$ \int_a^\infty f (x) \, dx $$ is well defined. If the appropriate limit exists, we attach the property "convergent" to that expression …

You will get the same answer because when you perform a change of variables, you change the limits of your integral as well (integrating in the complex plane requires defining a contour, of …