Question

What is `int e^(-x^2)dx`?

Answer 1

`int e^(-x^2) dx = sqrt(pi)/2 "erf"(x) + C`

`=sum_(n=0)^oo (-1)^n/((n!)(2n+1))x^(2n+1) + C`

Explanation:

The error function `"erf"(x)` is defined as follows:

`"erf"(x) = 2/sqrt(pi) int_0^x e^(-t^2) dt`

So `int e^(-x^2) dx = sqrt(pi)/2 "erf"(x) + C`

Can we find out the value of the integral without using this special non-elementary function?

We can at least express it in terms of a power series:

`e^t = sum_(n=0)^oo t^n/(n!)`

Substituting `t = -x^2` we get:

`e^(-x^2) = sum_(n=0)^oo (-1)^n/(n!)x^(2n)`

So:

`int e^(-x^2) dx = int (sum_(n=0)^oo (-1)^n/(n!)x^(2n)) dx`

`=sum_(n=0)^oo (-1)^n/((n!)(2n+1))x^(2n+1) + C`

Hence the power series for `"erf"(x)` is simply given by:

`"erf"(x) = 2/sqrt(pi) sum_(n=0)^oo (-1)^n/((n!)(2n+1))x^(2n+1)`

since `"erf"(x)` is the integral from `0` to `x` and this sum is `0` when `x = 0`.