Question

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

• A. \( \sqrt{\pi} \)
• B. \( \pi \)
• C. \( 1 \)
• D. \( \dfrac{\sqrt{\pi}}{2} \)

Hint:

The Gaussian integral is a standard result in calculus and statistics: \[ \int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi} \] It is closely related to the Normal distribution.

Answer 1

The Correct Option is A

Concept:
The integral \[ \int_{-\infty}^{\infty} e^{-x^2}\,dx \] is known as the Gaussian integral. It plays a fundamental role in probability theory, especially in the Normal distribution. A well-known result from calculus states: \[ \int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi} \] This result is usually derived by squaring the integral and converting it into polar coordinates.
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.
Using \(x^2 + y^2 = r^2\) and \(dx\,dy = r\,dr\,d\theta\): \[ I^2 = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} r\,dr\,d\theta \] \[ = 2\pi \int_{0}^{\infty} e^{-r^2} r\,dr \] Let \(u=r^2\), \(du=2r\,dr\): \[ \int_{0}^{\infty} e^{-r^2} r\,dr = \frac{1}{2} \] Thus, \[ I^2 = 2\pi \times \frac{1}{2} = \pi \]
Step 4: Take the square root.
\[ I = \sqrt{\pi} \] \[ \therefore \int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi} \]