If you have two continuous random variables (for example), X and Y, with joint pdf [itex]\rho_{X,Y}(x,y)[/itex], this can be written in terms of the conditional distribution [itex]\rho_{XY}(xy)[/itex] or [itex]\rho_{YX}(yx)[/itex] as:
[tex]\rho_{X,Y}(x,y) = \rho_{XY}(xy)\rho_Y(y) = \rho_{YX}(yx)\rho_X(x),[/tex]
where the individual distributions are
[tex]\rho_X(x) = \int_{\infty}^\infty dy~\rho_{X,Y}(x,y),[/tex]
and similarly for y. So, if you knew the distribution for y and you knew the condition distribution for x given y, you can calculate the distribution of x as
[tex]\rho_X(x) = \int_{\infty}^\infty dy~\rho_{XY}(xy)\rho_Y(y).[/tex]
Amusingly enough, I was dealing with this concept myself this week, although in the context of a much more intractable problem.
This said, I think you need to tweak your suggested problem. The variance of a distribution is positive, so it can't be normally distributed. If you use the standard deviation, I don't think you'll get a result, because you'll have a factor of [itex]1/\sigma[/itex] in your integration, why of course blows up at the origin, but the exponential will also be even in sigma, so the principal value might be zero.
