# Math Help - Integration: Error Function

1. ## Integration: Error Function

Hello,
I dont have the strongest math foundation, so please excuse me if this is a dumb question.
I was integrating the following in Maple:

$\int_0^1 e^{-x^2} dx$

$\frac{1}{2} \ erf(1) \ \sqrt{\pi}$

I've read that the erf is the error function but I don't really understand what it is. How can I obtain a 'Real' value for this?

2. Originally Posted by JavaJunkie
Hello,
I dont have the strongest math foundation, so please excuse me if this is a dumb question.
I was integrating the following in Maple:

$\int_0^1 e^{-x^2} dx$

$\frac{1}{2} \ erf(1) \ \sqrt{\pi}$

I've read that the erf is the error function but I don't really understand what it is. How can I obtain a 'Real' value for this?
By a little snake-oil (i.e. you don't care about the proof of why it's true right? If you do let me know_ we have that $\int_0^1e^{-x^2}dx=\sum_{n=0}^{\infty}\int_0^1 \frac{(-1)^nx^{2n}}{n!}dx=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)n!}$

3. Well, I would like to know why its true even though I don't have to.

$\int_0^1e^{-x^2}dx=\sum_{n=0}^{\infty}\int_0^1 \frac{(-1)^nx^{2n}}{n!}dx=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)n!}$ $\ \ \ \Longrightarrow$ thats a fourier series, isn't it?

4. Originally Posted by JavaJunkie
Well, I would like to know why its true even though I don't have to.

$\int_0^1e^{-x^2}dx=\sum_{n=0}^{\infty}\int_0^1 \frac{(-1)^nx^{2n}}{n!}dx=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)n!}$ $\ \ \ \Longrightarrow$ thats a fourier series, isn't it?
Not quite! But good guess. It comes from a Maclaurin series. The difficult part comes from why the following step is justified $\int_0^1\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!}=\sum_{n=0}^{\infty}\int_0^1\frac{(-1)^n x^{2n}}{n!}dx$. The idea is that $\left|\frac{(-1)^nx^{2n}}{n!}\right|\leqslant\frac{1}{n!}$ and so by the Weierstrass M-test we may conclude that the sum is uniformly convergent and thus the interchangeability of the series and the integral is justified.