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Let $X$ be a normal distribution with mean $0$ and variance $\sigma^2$. Then the Laplace transform of $X$ should be

$\int_0^\infty e^{-tx} \frac{1}{\sqrt{2\pi\sigma^2}}\exp\{-\frac{x^2}{2\sigma^2}\}dx$

Then I try to do integral on

$\int_0^\infty \exp\{-\frac{x^2}{2\sigma^2}-tx\}dx=\int_0^\infty \exp\{-\frac{1}{2\sigma^2}[x^2+2\sigma^2tx]\}dx$

finishing the square we can get

$\int_0^\infty \exp\{-\frac{1}{2\sigma^2}(x+\sigma^2 t)^2\}dx$

Then I replace $x+\sigma^2 t$ by $u$

$\int_{\sigma^2t}^\infty \exp\{-\frac{1}{2\sigma^2}u^2\}du$

Then I am stuck at this step.

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  • $\begingroup$ This is as far as it goes: the answer is in terms of the cumulative normal distribution function. $\endgroup$ Sep 18 '17 at 15:53
  • $\begingroup$ finishing the square yields another gaussian $\endgroup$
    – alain
    Jan 22 '18 at 10:51
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Well, in the general case you're looking at:

$$\mathscr{I}_{\space\text{n}}\left(\text{m}\space,\space\text{s}\right):=\int_0^\text{m}\frac{\text{n}}{\exp\left(\text{s}\cdot x+\pi\cdot\text{n}^2\cdot x^2\right)}\space\text{d}x\tag1$$

Substitute:

$$\text{u}:=\frac{2\cdot\pi\cdot\text{n}^2\cdot x+\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}\tag2$$

So, for the integral we get:

$$\mathscr{I}_{\space\text{n}}\left(\text{m}\space,\space\text{s}\right):=\text{n}\cdot\frac{\exp\left(\frac{\text{s}^2}{4\cdot\pi\cdot\text{n}^2}\right)}{2\cdot\text{n}}\int_{\frac{\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}}^{\frac{2\cdot\pi\cdot\text{n}^2\cdot\text{m}+\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}}\frac{2\cdot\exp\left(-\text{u}^2\right)}{\sqrt{\pi}}\space\text{d}\text{u}\tag2$$

This is a special integral:

$$\int_{\frac{\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}}^{\frac{2\cdot\pi\cdot\text{n}^2\cdot\text{m}+\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}}\frac{2\cdot\exp\left(-\text{u}^2\right)}{\sqrt{\pi}}\space\text{d}\text{u}=\left[\text{erf}\left(\text{u}\right)\right]_{\frac{\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}}^{\frac{2\cdot\pi\cdot\text{n}^2\cdot\text{m}+\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}}=$$ $$\text{erf}\left(\frac{2\cdot\pi\cdot\text{n}^2\cdot\text{m}+\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}\right)-\text{erf}\left(\frac{\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}\right)\tag3$$

So, we end up with:

$$\mathscr{I}_{\space\text{n}}\left(\text{m}\space,\space\text{s}\right)=\text{n}\cdot\frac{\exp\left(\frac{\text{s}^2}{4\cdot\pi\cdot\text{n}^2}\right)}{2\cdot\text{n}}\cdot\left\{\text{erf}\left(\frac{2\cdot\pi\cdot\text{n}^2\cdot\text{m}+\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}\right)-\text{erf}\left(\frac{\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}\right)\right\}\tag4$$

Now, your task is to simplify and to take:

$$\lim_{\text{m}\to\infty}\mathscr{I}_{\space\text{n}}\left(\text{m}\space,\space\text{s}\right)\tag5$$


Try to prove:

$$\lim_{\text{m}\to\infty}\text{erf}\left(\frac{2\cdot\pi\cdot\text{n}^2\cdot\text{m}+\text{s}}{2\cdot\sqrt{\pi}\cdot\text{n}}\right)=1\tag6$$

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