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does anyone know what is the Laplace transformation of the error function,

$\mathrm{erf}((X-a)/b)$ where $a$ and $b$ are constants?

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We have $$\mathrm{erf}(x) = \frac{2}{\sqrt{\pi}}\int_0^x e^{-x^2}\, \mathrm{d}x$$ You can use the properties of the Laplace transform and the fact that $\mathrm{erf}(0) = 0$ to show that: $$\mathcal{L}\{\mathrm{erf}(x)\} = \frac{2}{\sqrt{\pi}}\frac{G(s)}{s}$$ where $G(s) = \mathcal{L}\{e^{-x^2}\}$. Then: $$G(s) = \int_0^\infty e^{-x^2 - sx}\, \mathrm{d}x$$ Let $y = x + s/2$. Thus $y^2 = x^2 + sx + s^2/4$ and $x^2 + sx = y^2 - s^2/4$. Furthermore $\mathrm{d}y = \mathrm{d}x$, $x = 0 \Rightarrow y = s/2$, and $x \to \infty \Rightarrow y \to \infty$. Substituting yields: $$G(s) = \int_{s/2}^\infty e^{s^2/4} e^{-y^2}\, \mathrm{d}y = e^{s^2/4}\int_{s/2}^\infty e^{-y^2}\, \mathrm{d}y = e^{s^2/4}\frac{\sqrt{\pi}}{2}\mathrm{erfc}(s/2)$$ Now you can just substitute in $\mathcal{L}\{\mathrm{erf}(x)\}$: $$\mathcal{L}\{\mathrm{erf}(x)\} = \frac{e^{s^2/4}\mathrm{erfc}\left( \frac{s}{2} \right)}{s}$$ Then just use the Laplace properties for the constants $a$ and $b$.

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  • $\begingroup$ @userI10971 I found that $\mathrm{L}(f(X/b))=b \mathrm{ F(bs)}$ and $\mathrm{L}(f(X-a))=e^{-sa} F(s)$ . Am I supposed to think that $L(\mathrm{erf}((X-a)/b))=be^{-sa} \mathrm{ erf(bs)}$ ? $\endgroup$
    – Geotas
    Apr 5, 2017 at 11:41
  • $\begingroup$ @userI1097 is my previous comment correct? $\endgroup$
    – Geotas
    Apr 6, 2017 at 10:37
  • $\begingroup$ @Geotas it depends if erf is zero for $x<0$. If it is not, you cannot use the time shifting property. $\endgroup$
    – user110971
    Apr 6, 2017 at 12:31
  • $\begingroup$ @userI10971 erf is zero for x<0. In that case, is my first comment correct? $\endgroup$
    – Geotas
    Apr 10, 2017 at 11:36
  • $\begingroup$ @Geotas yes, if you use the Laplace transform of erf in your last line instead of erf. But erf has non zero values for all x < 0 $\endgroup$
    – user110971
    Apr 10, 2017 at 12:48

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