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.

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 …

@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, …

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 …

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.

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 …

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 …

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 $ …