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 and use …

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 …

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.

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

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 …

@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 …

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 …

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 …

5 An integral domain is a ring with no zero divisors, i.e. xy = 0 ⇒ x = 0 or y = 0. x y = 0 ⇒ x = 0 o r y = 0 Additionally it is a widespread convention to disallow as a domain the trivial one-element ring (or, …

Aug 20, 2023 · it works out because the integral is absolutely convergent (for x going to infinity the absolute value of the integrand behaves like x^2/x^4=1/x^2 which is integrable-look up cauchy …