one and two dimensional Gaussian integral

Question

a) Evaluate the one-dimensional Gaussian integral

$I(a)$ = $\int_R exp(-ax^2)dx$, $a>0$

b) evaluate the two-dimensional Gaussian integral using a)

$I_2(a,b)$ = $\int_{R^2} exp(-ax^2 -by^2)dxdy, a,b>0$

For a) I have done the following:

$\int_{-\infty}^{+\infty}$ $e^{-ax^2}dx = 2\int_0^{\infty}e^{-ax^2}dx $

$I^2$ = 4$\int_0^\infty$$\int_0^\infty$$e^{-a(x^2 + y^2)}dydx$

...

$I^2$ = $\sqrt{\pi a}$

$I$ = $\pi$$\sqrt a$

I am having difficulties understanding how to solve b) any help or guide will be appreciated.

Answer 1

Note that $e^{-ax^2-by^2} = e^{-a x^2}e^{-b y^2}$.

Then, note that $\iint_{\mathbb{R}^2} e^{-ax^2-by^2} dy dx = \iint_{\mathbb{R}^2} e^{-a x^2}e^{-b y^2} dy dx = \int_{\mathbb{R}} e^{-a x^2} dx \int_{\mathbb{R}} e^{-b y^2} dy$, and apply your result from part (a).

Also, note $\int_{\mathbb{R}} \frac{1}{\sqrt{2 \pi \sigma}} e^{-\frac{x^2}{2 \sigma^2}} dx = 1$ (compare this to your answer for part (a)).

Answer 2

Note the suggested answer in the original question is wrong. Should be: $$I = \sqrt{\frac{\pi}{a}}$$