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.

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 …

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 …

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...

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

You will get the same answer because when you perform a change of variables, you change the limits of your integral as well (integrating in the complex plane requires defining a contour, of course, so you'll …

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)$ …