The integral which you describe has no closed form which is to say that it cannot be expressed in elementary functions. For example, you can express $\int x^2 \mathrm {d}x$ in elementary …

@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, …

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.

Nov 29, 2013 · Using "indefinite integral" to mean "antiderivative" (which is unfortunately common) obscures the fact that integration and anti-differentiation really are different things in general.

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

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 …

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

For an integral of the form $$\tag {1}\int_a^ {g (x)} f (t)\,dt,$$ you would find the derivative using the chain rule. As stated above, the basic differentiation rule for integrals is:

Feb 17, 2025 · So an improper integral is a limit which is a number. Does it make sense to talk about a number being convergent/divergent? It's fixed and does not change with respect to the …

Dec 15, 2017 · A different type of integral, if you want to call it an integral, is a "path integral". These are actually defined by a "normal" integral (such as a Riemann integral), but path …