I think you mean
[math]G(x) = \dfrac{1}{ \pi } e^{-(x- x_m)^2/ ( \Delta x )^2}[/math]
[math]G(x, y) = \dfrac{1}{ \pi ^2 } e^{-(x- x_m)^2/ ( \Delta x )^2} ~ e^{-(y- y_m)^2/ ( \Delta y )^2}[/math]
This is a bit different from how the Gaussian function is usually written but you can fix that if you need to. Any Calculus methods will work the same way.
Take G(x, y) and modify it a bit, just to show the method.
[math]G(x, y) = A e^{-x^2} e^{-y^2}[/math]
[math]\int_{- \infty }^{ \infty } \int_{- \infty }^{ \infty } A e^{-x^2 - y^2 } ~ dx ~ dy[/math]
Convert to polar coordinates:
[math]= \int_{0}^{ \infty } \int_{0}^{2 \pi } A e^{-r^2} ~ r ~ dr ~ d \theta [/math]
You can now use integration by parts.
The G(x) integration is done in exactly the same way. [math]G(x) G(y) = G(x, y)[/math], so [math]\int G(x) ~ dx = \sqrt{ \int G(x, y) dx dy}[/math]
-Dan