(Why √
2π?) Use polar coordinates to show that
Z
R
Z
R
e
−x
2
e
−y
2
dxdy = 2π.
Conclude that
1
√
2π
Z
R
e
−x
2
dx = 1.

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.

Use polar coordinates to show that \( \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2} e^{-y^2} \, dx \, dy = 2\pi \). Conclude that \( \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} e^{-x^2} \, dx = 1 \).

Learn how to use polar coordinates to evaluate the Gaussian integral \( \int_{-\infty}^{\infty} e^{-x^2} \, dx \) and show that it equals \( \sqrt{\pi} \). This tutorial includes a detailed step-by-step solution and covers concepts like double integrals, polar coordinates, and Gaussian functions, suitable for calculus students.

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