Double Integration

**wantanswers** · June 30th 2008, 07:38 AM

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

**wingless** · June 30th 2008, 08:11 AM

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

**ThePerfectHacker** · June 30th 2008, 10:15 AM

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 $2\pi$ ).

$\int_{-\infty}^{\infty} \int_{-\infty}^{z} e^{-m^2/2} e^{-kz}e^{-z^2/2} dz ~ dm$

**CaptainBlack** · June 30th 2008, 11:13 PM

Originally Posted by wantanswers

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:

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

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

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

RonL

**wantanswers** · July 7th 2008, 06:13 AM

This is the integral ,

which should give me

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

**mr fantastic** · July 7th 2008, 06:32 PM

Originally Posted by wantanswers

This is the integral ,

which should give me

$ 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.

**wantanswers** · July 8th 2008, 05:22 AM

This is the original problem

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

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

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

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

$\Phi(y)=\int_{-\infty}^{y} e^{-m^2/2} dm$

**mr fantastic** · July 11th 2008, 06:54 AM

Originally Posted by wantanswers

This is the original problem

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

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

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

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

$\Phi(y)=\int_{-\infty}^{y} e^{-m^2/2} dm$

Is this the original question, or your mathematical interpretation of it? If the former, I'm afraid I don't see a solution at this point in time. Has it come from a statistics question?

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