Question

What is the integral of the error function?

Answer 1

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

Explanation:

We will use the definition of the error function:

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

Along with integration by substitution, integration by parts, and the fundamental theorem of calculus.

Integration by Parts:

Let `u = "erf"(x)` and `dv = dt`

Then, by the fundamental theorem of calculus,

`du = 2/sqrt(pi)e^(-x^2)` and `v = x`

By the integration by parts formula `intudv = uv - intvdu`

`int"erf"(x)dx = x"erf"(x) - int2/sqrt(pi)xe^(-x^2)dx`

Integration by Substitution:

To evaluate the remaining integral, let `u = -x^2`

Then `du = -2xdx` and so

`-int2/sqrt(pi)xe^(-x^2)dx = 1/sqrt(pi)inte^udu`

`=1/sqrt(pi)e^u + C`

`=e^(-x^2)/sqrt(pi)+C`

Putting it all together, we get our final result:

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