# Laplace Transform of Error Function

## Theorem

$\laptrans {\map \erf t} = \dfrac 1 s \map \exp {\dfrac {s^2} 4} \map \erfc {\dfrac s 2}$

where:

$\laptrans f$ denotes the Laplace transform of the function $f$
$\erf$ denotes the error function
$\erfc$ denotes the complementary error function
$\exp$ denotes the exponential function.

## Proof

By Derivative of Error Function, we have:

$\ds \map {\frac \d {\d t} } {\map \erf t} = \frac 2 {\sqrt \pi} e^{-t^2}$

By Primitive of Exponential Function, we have:

$\ds \int e^{-s t} \rd t = -\frac {e^{-s t} } s$

So:

 $\ds \laptrans {\map \erf t}$ $=$ $\ds \int_0^\infty e^{-s t} \map \erf t \rd t$ Definition of Laplace Transform $\ds$ $=$ $\ds \intlimits {-\frac 1 s e^{-s t} \map \erf t} 0 \infty - \int_0^\infty \paren {-\frac 2 {\sqrt \pi} \frac {e^{-s t} } s e^{-t^2} } \rd t$ Integration by Parts $\ds$ $=$ $\ds -\frac 1 s \lim_{t \mathop \to \infty} \paren {e^{-s t} \map \erf t} + \frac 1 s e^0 \erf 0 + \frac 2 {s \sqrt \pi} \int_0^\infty \exp \paren {-s t - t^2} \rd t$

We have:

 $\ds \lim_{t \mathop \to \infty} \paren {e^{-s t} \map \erf t}$ $=$ $\ds \paren {\lim_{t \mathop \to \infty} e^{-s t} } \paren {\lim_{t \mathop \to \infty} \map \erf t}$ Product Rule for Limits of Real Functions $\ds$ $=$ $\ds 0 \times 1$ Exponential Tends to Zero and Infinity, Limit to Infinity of Error Function $\ds$ $=$ $\ds 0$

We also have:

 $\ds \frac 1 s e^0 \erf 0$ $=$ $\ds \frac 1 s \int_0^0 e^{-t^2} \rd t$ Exponential of Zero, Definition of Error Function $\ds$ $=$ $\ds 0$ Definite Integral on Zero Interval

Therefore:

 $\ds \laptrans {\map \erf t}$ $=$ $\ds \frac 2 {s \sqrt \pi} \int_0^\infty \map \exp {-\paren {t^2 + s t} } \rd t$ $\ds$ $=$ $\ds \frac 2 {s \sqrt \pi} \int_0^\infty \map \exp {-\paren {\paren {t + \frac s 2}^2 - \frac {s^2} 4} } \rd t$ completing the square $\ds$ $=$ $\ds \frac 2 {s \sqrt \pi} \map \exp {\frac {s^2} 4} \int_0^\infty \map \exp {-\paren {t + \frac s 2}^2} \rd t$ Exponential of Sum $\ds$ $=$ $\ds \frac 2 {s \sqrt \pi} \map \exp {\frac {s^2} 4} \int_{\frac s 2}^\infty \map \exp {-u^2} \rd u$ substituting $u = t + \dfrac s 2$ $\ds$ $=$ $\ds \frac 1 s \map \exp {\frac {s^2} 4} \paren {\frac 2 {\sqrt \pi} \int_{\frac s 2}^\infty \map \exp {-u^2} \rd u}$ $\ds$ $=$ $\ds \frac 1 s \map \exp {\frac {s^2} 4} \map \erfc {\frac s 2}$ Definition of Complementary Error Function

$\blacksquare$