Can you help with this integration

int (inf,-inf)int (z,-inf)( exp-m(square)/2)dm(2Π) exp(-kz) ( exp-z(square)/2)dz

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- Jun 30th 2008, 07:38 AMwantanswersDouble Integration
Can you help with this integration

int (inf,-inf)int (z,-inf)( exp-m(square)/2)dm(2Π) exp(-kz) ( exp-z(square)/2)dz - Jun 30th 2008, 08:11 AMwingless
It looks divergent. Are you sure you wrote this correctly? Can you rewrite it using more parentheses or LaTeX?

- Jun 30th 2008, 10:15 AMThePerfectHacker
- Jun 30th 2008, 11:13 PMCaptainBlack
In LaTeX:

$\displaystyle \int^{\infty}_{-\infty} \left( \int^z_{-\infty} e^{-m^2/2}\ dm \right) (2 \pi) e^{-kz} e^{-z^2/2}\ dz$

Which is almost certainly non-elementary, as the integral in the brackets is a error function, or $\displaystyle \sqrt{2 \pi}\Phi(x)$, where $\displaystyle \Phi(x)$ is the cumulative standard normal distribution.

So the question is why do you want this and would a numerical value do?

RonL - Jul 7th 2008, 06:13 AMwantanswers
This is the integral ,

http://www.mathhelpforum.com/math-he...966fd0ea-1.gif

which should give me

$\displaystyle

e^{k^2/2}\int_{-\infty}^{-k/\surd2} e^{-m^2/2} dm

$ - Jul 7th 2008, 06:32 PMmr fantastic
- Jul 8th 2008, 05:22 AMwantanswers
This is the original problem

$\displaystyle

\int_{\epsilon+\alpha/k}^{\infty} xF(x)f(x)dx

$

where F(x)=$\displaystyle \Phi(y)$

f(x)=$\displaystyle (2\Pi )^{-1/2} e^{-ky}e^{-y^2/2}$

y=$\displaystyle -k ^{-1}log [1-k(x-\epsilon)/\alpha]$

$\displaystyle \Phi(y)=\int_{-\infty}^{y} e^{-m^2/2} dm$ - Jul 11th 2008, 06:54 AMmr fantastic