# Double Integration

• Jun 30th 2008, 07:38 AM
Double 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 AM
wingless
It looks divergent. Are you sure you wrote this correctly? Can you rewrite it using more parentheses or LaTeX?
• Jun 30th 2008, 10:15 AM
ThePerfectHacker
Quote:

Originally Posted by wingless
It looks divergent. Are you sure you wrote this correctly? Can you rewrite it using more parentheses or LaTeX?

How about this? (Ignore $\displaystyle 2\pi$).

$\displaystyle \int_{-\infty}^{\infty} \int_{-\infty}^{z} e^{-m^2/2} e^{-kz}e^{-z^2/2} dz ~ dm$
• Jun 30th 2008, 11:13 PM
CaptainBlack
Quote:

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

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 AM
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 PM
mr fantastic
Quote:

This is the integral ,

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

which should give me

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

Why should it?

If you state the original problem that has led to this question, some progress might be made.
• Jul 8th 2008, 05:22 AM
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 AM
mr fantastic
Quote:

$\displaystyle \int_{\epsilon+\alpha/k}^{\infty} xF(x)f(x)dx$
where F(x)=$\displaystyle \Phi(y)$
f(x)=$\displaystyle (2\Pi )^{-1/2} e^{-ky}e^{-y^2/2}$
y=$\displaystyle -k ^{-1}log [1-k(x-\epsilon)/\alpha]$
$\displaystyle \Phi(y)=\int_{-\infty}^{y} e^{-m^2/2} dm$