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Is it possible to calculate the Laplace Transform of a normal cumulative distribution?

$\int_{0}^{\infty}{ \left(\int_{-\infty}^{x} {1\over \sqrt(2\pi)}e^{-y^2\over2} \mathrm dy\right)} \mathrm e^{-xs} dx=I\tag1$

dleal
  • 271
  • You probably meant the bilateral Laplace transform. For $s \in \mathbb{R}$ and by analytic continuation for $s \in \mathbb{C}$ : $\int_{-\infty}^\infty e^{-x^2/2} e^{-sx}dx = e^{s^2/2}\int_{-\infty}^\infty e^{-(x+s)^2/2} dx= e^{s^2/2}\int_{-\infty}^\infty e^{-x^2/2} dx=e^{s^2/2} \sqrt{2\pi}$.

    Therefore for $\Re(s) > 0$ : $\int_{-\infty}^\infty (\int_{-\infty}^x e^{-y^2/2}dy) e^{-sx}dx = -\int_{-\infty}^\infty e^{-x^2/2} \frac{e^{-sx}}{-s}dx = \frac{1}{s}e^{s^2/2} \sqrt{2\pi}$

     – reuns Sep 08 '17 at 09:55
  • I probably don't know enough, but i have a non-homogeneous differential equation, where the normal density function enters into the non-homogeneous part. Should I apply the bilateral or the regular Laplace transform? – dleal Sep 08 '17 at 18:31

2 Answers2

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Well, we have:

$$\mathscr{I}\left(\text{s}\right)=\mathscr{L}_x\left[\int_{-\infty}^x\frac{\exp\left(-\frac{\text{y}^2}{2}\right)}{\sqrt{2\pi}}\space\text{d}\text{y}\right]_{\left(\text{s}\right)}:=\int_0^\infty\left\{\int_{-\infty}^x\frac{\exp\left(-\frac{\text{y}^2}{2}\right)}{\sqrt{2\pi}}\space\text{d}\text{y}\right\}\cdot e^{-\text{s}x}\space\text{d}x=$$ $$\frac{1}{\sqrt{2\pi}}\int_0^\infty\exp\left(-\text{s}\cdot x\right)\int_{-\infty}^x\exp\left(-\frac{\text{y}^2}{2}\right)\space\text{d}\text{y}\space\text{d}x\tag1$$

For the inner integral, substitute $\text{u}:=\frac{\text{y}}{\sqrt{2}}$:

$$\int_{-\infty}^x\exp\left(-\frac{\text{y}^2}{2}\right)\space\text{d}\text{y}=\frac{\sqrt{\pi}}{\sqrt{2}}\int_{-\infty}^\frac{x}{\sqrt{2}}\frac{2\cdot\exp\left(-\text{u}^2\right)}{\sqrt{\pi}}\space\text{d}\text{u}=$$ $$\frac{\sqrt{\pi}}{\sqrt{2}}\cdot\lim_{\text{n}\to-\infty}\space\left[\text{erf}\left(\text{u}\right)\right]_\text{n}^\frac{x}{\sqrt{2}}=\frac{\sqrt{\pi}}{\sqrt{2}}\cdot\left(\text{erf}\left(\frac{x}{\sqrt{2}}\right)-\lim_{\text{n}\to-\infty}\space\text{erf}\left(\text{n}\right)\right)=$$ $$\frac{\sqrt{\pi}}{\sqrt{2}}\cdot\left(\text{erf}\left(\frac{x}{\sqrt{2}}\right)-\left(-1\right)\right)=\frac{\sqrt{\pi}}{\sqrt{2}}\cdot\left(1+\text{erf}\left(\frac{x}{\sqrt{2}}\right)\right)\tag2$$

So, we get:

$$\mathscr{I}\left(\text{s}\right)=\frac{1}{\sqrt{2\pi}}\int_0^\infty\exp\left(-\text{s}\cdot x\right)\cdot\frac{\sqrt{\pi}}{\sqrt{2}}\cdot\left(1+\text{erf}\left(\frac{x}{\sqrt{2}}\right)\right)\space\text{d}x=$$ $$\frac{1}{2}\cdot\left\{\int_0^\infty\exp\left(-\text{s}\cdot x\right)\cdot1\space\text{d}x+\int_0^\infty\exp\left(-\text{s}\cdot x\right)\cdot\text{erf}\left(\frac{x}{\sqrt{2}}\right)\space\text{d}x\right\}=$$ $$\frac{1}{2}\cdot\left\{\mathscr{L}_x\left[1\right]_{\left(\text{s}\right)}+\mathscr{L}_x\left[\text{erf}\left(\frac{x}{\sqrt{2}}\right)\right]_{\left(\text{s}\right)}\right\}=\frac{1}{2}\cdot\left\{\frac{1}{\text{s}}+\mathscr{L}_x\left[\text{erf}\left(\frac{x}{\sqrt{2}}\right)\right]_{\left(\text{s}\right)}\right\}\tag3$$

Jan Eerland
  • 28,671
  • The bilateral Laplace transform has a closed-form. For the unilateral one we need $\text{erf}(s)$. For $s \in \mathbb{R}$ and by analytic continuation for $s \in \mathbb{C}$ : $\int_0^\infty e^{-x^2/2} e^{-sx}dx = e^{s^2/2}\int_0^\infty e^{-(x+s)^2/2} dx= e^{s^2/2}\int_{s}^\infty e^{-x^2/2} dx=e^{s^2/2} \text{erf}(-s)$.

    Therefore for $\Re(s) > 0$ : $\int_0^\infty (\int_{-\infty}^x e^{-y^2/2}dy) e^{-sx}dx = \frac{\text{erf}(0)}{-s} -\int_0^\infty e^{-x^2/2} \frac{e^{-sx}}{-s}dx = \frac{\sqrt{\pi/2}}{-s}+\frac{1}{s}e^{s^2/2}\text{erf}(-s)$

     – reuns Sep 08 '17 at 10:00
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I think the best way would be to note that

$\Phi(x) = \frac{1 + erf(x/\sqrt{2}))}{2} $

and find theLT of the error function which is well known

dleal
  • 271
  • See my two comments ? And the bilateral Laplace transform allows to solve $\sum_{k = 0}^K c_k f^{(k)} = \Phi$ easily. – reuns Sep 10 '17 at 19:49
  • Thank you, I noticed that in one of your comments you used the erf. I would be glad to accept that as an answer – dleal Sep 10 '17 at 19:54