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 …

4 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 integral being …

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 indefinite integral." One is free …

@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 …

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.

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 any function f …

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 …

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

Dec 1, 2024 · In calculus we've been introduced first with indefinite integral, then with the definite one. Then we've been introduced with the concept of double (definite) integral and multiple (definite) integ...

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 $ (2)$ …