why do we define $\int{e^{-t^2}dt}=\frac{\sqrt{\pi}}{2}erf(t)+c$?

Question

I am relearning differential equation on my own, and came across a problem that gives the integral as erf function. Why is it defined in this way $\int{e^{-t^2}dt}=\frac{\sqrt{\pi}}{2}erf(t)+c$? I couldn't integrate it, and I am guessing it's not possible to integrate so that's why it's defined?

Answer 1

Imagine you're in a world where the $ \log $ function doesn't exist. How do you integrate $ \int 1/t \, dt $?

Here's one way: we simply define a function called $ \operatorname{plog} $ such that

$$ \operatorname{plog}(x) := \int_{1}^{x} \frac{1}{t} \, dt. $$

Then $ \int 1/t \, dt = \operatorname{plog}(t) + C $ and we can continue as we would in the world where $ \log $ exists but instead we have $ \operatorname{plog} $.

How about $ e^{-t^2} $? You're correct, it turns out it cannot be integrated in terms of so called "elementary" functions (combinations of: powers, roots, $\exp, \log, \sin, \cos, \sin^{-1}, \cos^{-1}, \ldots$). This can apparently be shown using the Risch Algorithm.

We therefore define a function called the error function by

$$ \operatorname{erf}(x) := \frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^2} \, dt, $$

so that, as you say, $\int{e^{-t^2}dt}=\frac{\sqrt{\pi}}{2}\operatorname{erf}(t)+C$. Then $ \operatorname{erf} $ can't be expressed in terms elementary functions, so we add it to our list of "non-elementary" functions which contains a bunch of other special functions that cannot be expressed in terms of the elementary functions. Examples include

• Exponential integral: $ \operatorname{Ei}(x) = \int_{-x}^{\infty} e^t/t \, dt$
• Logarithmic integral: $ \operatorname{li}(x) = \int_{0}^{x} 1/\log(t) \, dt$

Some non-elementary functions can't even be written in terms of an integral (these functions are called non-Liouvillian if that's the case). Examples of those are:

• Bessel functions: solutions to the differential equation $x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + \left(x^2 - \alpha^2 \right)y = 0$
• Hypergeometric functions: special types of infinite series

You might be wondering: what's with the funny $ \frac{2}{\sqrt{\pi}} $ out the front of the integral in the definition of $\operatorname{erf}$? That's a normalisation constant to ensure $ \operatorname{erf}(x) = \pm 1 $ as $x \to \pm \infty$, which is important to the statisticians and physicists, have a look here: Why is the error function defined as it is?

Answer 2

I many problems one can prove that a solution exists, for example which can be expressed on the form of an infinite series. But one fails to express it with a combination of a limited number of elementary functions.

In order to deal with closed form instead of infinite series or instead of other complicated expressions for example involving non elementary integrals, some new functions where defined so called "special functions".

This is the case of the erf function. To understand why this special function was defined and what is the advantage to use it in calculus, one have to acquire a general knownledge about the "special functions".

http://mathworld.wolfram.com/SpecialFunction.html

https://en.wikipedia.org/wiki/Special_functions

A paper for general public : https://fr.scribd.com/doc/14623310/Safari-on-the-country-of-the-Special-Functions-Safari-au-pays-des-fonctions-speciales