In article <1107884944.664326.49260@f14g2000cwb.googlegroups.com>,
Jimbo <jamesdanielturner@hotmail.com> wrote:

> Let
> 
> I = int(0,inf){(1-exp(-z))exp(-z/2)/z }dz

I think the question was exp(-z^2) not exp(-z/2)

> 
> and z^(-1) = int(0,inf){exp(-az)}da
> 
> So That
> 
> I = int(0,inf){ int(0,inf){(1-exp(-z))exp(-z/2)*exp(-az) }dz }da
> 
>   = int(0,inf){ int(0,inf){exp(-(a+1/2)z ) - exp(-(a+3/2)z )}dz }da
> 
>   = int(0,inf){ 1/(a+1/2) - 1/(a+3/2) } da
> 
>   = int(0,inf){ d/da( ln( (a+1/2)/(a+3/2) ) ) }da
> 
>   = ln( (a+1/2)/(a+3/2) )|a=>inf - ln( (a+1/2)/(a+3/2) )|a=>0
> 
>   = 0 - ln(1/3)
> 
>   = - ln(1/3) = 1.098162288668109....
> 
> How exact do you need!!!!
> 
> Jim Turner
> Manager, Business Development
> Dynacs Military and Defense
> 1300 Hercules, Suite 205
> Houston, TX 77058
> 281-226-5213
> jturner@dynacsmdi.com
>