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 …

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.

Mar 17, 2021 · For the second integral we note that $$\lim_ {x\to\infty}\frac {x^ {t-1}e^ {-x}} {\frac {1} {x^2}} = \lim_ {x\to\infty}x^ {t+1}e^ {-x} = 0$$ and again by comparison test, the integral $ …

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 …

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 …

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 …

@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, …

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

May 21, 2020 · How to evaluate the integral $\int e^ {x^3}dx $ Ask Question Asked 12 years, 10 months ago Modified 1 year, 5 months ago