Let's solve this step by step using polar coordinates.

Step 1: Set up the double integral

We are given the double integral:

$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2} e^{-y^2} \, dx \, dy$

This can be rewritten as:

$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2)} \, dx \, dy$

Step 2: Convert to polar coordinates

In polar coordinates, the variables $x$ and $y$ are replaced by:

$x = r \cos \theta, \quad y = r \sin \theta$

The corresponding area element $dx \, dy$ becomes:

$dx \, dy = r \, dr \, d\theta$

Also, in polar coordinates, $x^2 + y^2 = r^2$. So, the exponential term becomes $e^{-r^2}$.

Now the integral becomes:

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

Step 3: Solve the radial part

First, solve the radial integral:

$\int_0^\infty e^{-r^2} r \, dr$

To solve this, we use the substitution $u = r^2$, which gives $du = 2r \, dr$, or $r \, dr = \frac{du}{2}$. The integral becomes:

$\int_0^\infty e^{-u} \frac{du}{2} = \frac{1}{2} \int_0^\infty e^{-u} \, du = \frac{1}{2}$

Step 4: Solve the angular part

Now, solve the angular integral:

$\int_0^{2\pi} d\theta = 2\pi$

Step 5: Combine the results

Combining both integrals, we get:

$\int_0^{2\pi} \int_0^\infty e^{-r^2} r \, dr \, d\theta = 2\pi \times \frac{1}{2} = \pi$

Therefore:

$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2 + y^2)} \, dx \, dy = \pi$

Step 6: Conclude the 1D Gaussian integral

The integral of $e^{-x^2}$ over the entire real line is related to this result. We have:

$\left( \int_{-\infty}^\infty e^{-x^2} \, dx \right)^2 = \pi$

Taking the square root of both sides:

$\int_{-\infty}^\infty e^{-x^2} \, dx = \sqrt{\pi}$

Step 7: Normalize the Gaussian integral

Now we conclude that the normalized version of the Gaussian integral is:

$\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-x^2} \, dx = 1$

This shows that the integral of the Gaussian function with the proper normalization factor is equal to 1, which is fundamental in probability theory and statistics for the normal distribution.

---

Would you like more details or have any questions about the process?

Here are 5 related questions to explore further:

1. How do polar coordinates simplify double integrals?
2. Why is the substitution $u = r^2$ useful in radial integrals?
3. Can you derive the Gaussian integral for any general quadratic exponent?
4. What role does the factor $\frac{1}{\sqrt{2\pi}}$ play in normalizing distributions?
5. How does this relate to the standard normal distribution in probability?

Tip: When solving integrals involving exponentials in 2D, polar coordinates are often the simplest approach, especially for radially symmetric functions.