# Math Help - Double Integration

1. ## 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

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

3. 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 $2\pi$).

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

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

5. This is the integral ,

which should give me

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

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.

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

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

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?