
solving the integral of $e^ {x^2}$ - Mathematics Stack Exchange
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 functions …
What is the integral of 1/x? - Mathematics Stack Exchange
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.
What is an integral? - Mathematics Stack Exchange
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 integrals do not seek to …
calculus - Is there really no way to integrate $e^ {-x^2 ...
@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, move the …
A closed form for an integral involving arcsin and a square-root
Jan 7, 2026 · A closed form for an integral involving arcsin and a square-root Ask Question Asked 12 days ago Modified 11 days ago
An integral that Wolfram Alpha can compute, but I want to know the ...
Jan 13, 2026 · The integral is $$\int_0^ {\infty}\frac {e^ {-\frac {1+4y^2} {4y}}} {\sqrt {4\pi y}}dy$$ which Wolfram Alpha computes to $\frac 1 {2e}$. I would like to know the steps. I noticed the integral is …
Separating an integral - Mathematics Stack Exchange
Nov 4, 2023 · Separating an integral Ask Question Asked 2 years, 3 months ago Modified 2 years, 3 months ago
calculus - Evaluate an integral involving a series and product in the ...
Feb 6, 2026 · Evaluate an integral involving a series and product in the denominator Ask Question Asked 3 days ago Modified 2 days ago
Can the integral closure of a ring be taken intrinsically?
Oct 11, 2025 · However, one "intrinsic integral closure" that is often used is the normalization, which in the case on an integral domain is the integral closure in its field of fractions. It's the maximal integral …
How to calculate the integral in normal distribution?
If by integral you mean the cumulative distribution function $\Phi (x)$ mentioned in the comments by the OP, then your assertion is incorrect.