Primitive of Error Function

Theorem

$\ds \int \map \erf x \rd x = x \map \erf x + \frac 1 {\sqrt \pi} e^{-x^2} + C$

where $\erf$ denotes the error function.

Proof

By Derivative of Error Function, we have:

$\dfrac \d {\d x} \paren {\map \erf x} = \dfrac 2 {\sqrt \pi} e^{-x^2}$

So:

\(\ds \int \map \erf x \rd x\)

\(=\)

\(\ds \int 1 \times \map \erf x \rd x\)

\(\ds \)

\(=\)

\(\ds x \map \erf x - \frac 2 {\sqrt \pi} \int x e^{-x^2} \rd x\)

Integration by Parts

\(\ds \)

\(=\)

\(\ds x \map \erf x + \frac 1 {\sqrt \pi} \int \paren {-2 x e^{-x^2} } \rd x\)

\(\ds \)

\(=\)

\(\ds x \map \erf x + \frac 1 {\sqrt \pi} \int e^u \rd u\)

substituting $u = -x^2$

\(\ds \)

\(=\)

\(\ds x \map \erf x + \frac 1 {\sqrt \pi} e^u + C\)

Primitive of Exponential Function

\(\ds \)

\(=\)

\(\ds x \map \erf x + \frac 1 {\sqrt \pi} e^{-x^2} + C\)

substituting back for $u$

$\blacksquare$