Laplace transform of normal cumulative

Question

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

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

Answer 1

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:

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

Answer 2

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