Solving 2D Gaussian Integral w.r.t x and y and also w.r.t polar coordinates

w.r.t x and y

To find the double integral of $e^{-\frac{x^2}{2\sigma^2} - \frac{y^2}{2\sigma^2}}$ with respect to $x$ and $y$, you need to specify the limits of integration. Since the function involves the Gaussian distribution, it's common to integrate over the entire real line. The result is related to the normalization constant of the 2D Gaussian distribution.

The double integral is given by:

$\iint e^{-\frac{x^2}{2\sigma^2} - \frac{y^2}{2\sigma^2}} \,dx \,dy$

Assuming integration over the entire real line, this double integral can be evaluated, and the result is:

$\iint e^{-\frac{x^2}{2\sigma^2} - \frac{y^2}{2\sigma^2}} \,dx \,dy = 2\pi\sigma^2$

This result is obtained by expressing the double integral as the product of two Gaussian integrals (one for $x$ and one for $y$), each contributing a factor of $\sqrt{2\pi\sigma^2}$, and then multiplying those factors together.

Keep in mind that the specific limits of integration and the context in which this double integral is applied may vary, so make sure to adjust accordingly based on the problem at hand.

w.r.t. polar coordinates

When changing variables in a double integral from Cartesian to polar coordinates, the area element dx dy transforms to r dr dθ. This is a crucial step in evaluating integrals over regions that are more naturally expressed in polar coordinates, such as circles and sectors.

The 2D Gaussian integral involves integrating a Gaussian function over two variables, typically $x$ and $y$. The general form of the 2D Gaussian integral is:

$\iint e^{-\frac{x^2 + y^2}{2\sigma^2}} \,dx \,dy$

Here, $\sigma$ is a parameter representing the standard deviation of the Gaussian distribution. The limits of integration would depend on the specific context of the problem, but commonly, the integration is performed over the entire real line for both $x$ and $y$.

The 2D Gaussian integral can be evaluated by expressing it in polar coordinates. The transformation from Cartesian to polar coordinates is given by $x = r\cos(\theta)$ and $y = r\sin(\theta)$, where $r$ is the radius and $\theta$ is the angle.

The integral then becomes:

$\int_0^{2\pi} \int_0^\infty e^{-\frac{r^2}{2\sigma^2}} \,r \,dr \,d\theta$

This integral simplifies to:

$2\pi \sigma^2$

This result is obtained by recognizing that the radial part of the integral is a standard Gaussian integral, and the angular part yields a factor of $2\pi$.

So, the 2D Gaussian integral evaluates to $2\pi\sigma^2$, which is the area under the 2D Gaussian distribution.