Question

What is the value of the integral \( \int_{-\infty}^{\infty} e^{-x^2} \, dx \)?

• A. \(0\)
• B. \(1\)
• C. \(\sqrt{\pi}\)
• D. \(\pi\)

Hint:

\[ \int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi} \] A very important standard result in probability and analysis.

Answer 1

The Correct Option is C

Concept: Gaussian Integral
The integral: \[ \int_{-\infty}^{\infty} e^{-x^2} dx \] is a standard result known as the Gaussian integral.
Step 1: Define the integral
Let: \[ I = \int_{-\infty}^{\infty} e^{-x^2} dx \]
Step 2: Square the integral
\[ I^2 = \left(\int_{-\infty}^{\infty} e^{-x^2} dx \right)\left(\int_{-\infty}^{\infty} e^{-y^2} dy \right) \] \[ = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+y^2)} dx\,dy \]
Step 3: Convert to polar coordinates
\[ x^2 + y^2 = r^2, \quad dx\,dy = r\,dr\,d\theta \] \[ I^2 = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} r\,dr\,d\theta \]
Step 4: Evaluate the integral
\[ \int_{0}^{\infty} e^{-r^2} r\,dr = \frac{1}{2} \] \[ I^2 = 2\pi \cdot \frac{1}{2} = \pi \]
Step 5: Final result
\[ I = \sqrt{\pi} \]