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:


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

By Primitive of Exponential Function, we have:


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

So:

We have:

We also have:

Therefore: