Question

The value of $\int_0^{\infty} \int_0^{\infty} e^{-(x^2+y^2)} dx dy$ is}

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{6}$
• C. $\frac{\sqrt{\pi}}{2}$
• D. $\frac{\pi}{4}$

Hint:

The standard Gaussian integral $\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$ is extremely useful.
By symmetry, over the interval $[0, \infty)$, it is $\frac{\sqrt{\pi}}{2}$.
The product of two such independent integrals is $(\frac{\sqrt{\pi}}{2})^2 = \frac{\pi}{4}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to evaluate the double integral of $e^{-(x^2+y^2)}$ over the first quadrant ($0 \le x < \infty$ and $0 \le y < \infty$).

Step 2: Key Formula or Approach:
Since the limits of integration are independent and the integrand can be factored into a product of functions of $x$ and $y$, we can separate the double integral into a product of two single integrals:
\[ I = \int_0^{\infty} \int_0^{\infty} e^{-(x^2+y^2)} dx dy = \left(\int_0^{\infty} e^{-x^2} dx\right) \left(\int_0^{\infty} e^{-y^2} dy\right) \] We can evaluate this using the standard Gaussian integral value $\int_0^{\infty} e^{-u^2} du = \frac{\sqrt{\pi}}{2}$.

Step 3: Detailed Explanation:

• Let us evaluate each single integral:
\[ \int_0^{\infty} e^{-x^2} dx = \frac{\sqrt{\pi}}{2} \] \[ \int_0^{\infty} e^{-y^2} dy = \frac{\sqrt{\pi}}{2} \]
• Multiply the two independent results:
\[ I = \left(\frac{\sqrt{\pi}}{2}\right) \cdot \left(\frac{\sqrt{\pi}}{2}\right) = \frac{\pi}{4} \]
• Alternatively, we can evaluate this by transforming the integral into polar coordinates ($x = r\cos\theta$, $y = r\sin\theta$, and $dx dy = r dr d\theta$):
- The limits for the first quadrant are $0 \le r < \infty$ and $0 \le \theta \le \frac{\pi}{2}$.
\[ I = \int_0^{\pi/2} d\theta \int_0^{\infty} e^{-r^2} r dr \] - Integrating with respect to $\theta$:
\[ \int_0^{\pi/2} d\theta = \frac{\pi}{2} \] - Integrating with respect to $r$ using the substitution $u = r^2 \implies du = 2r dr$:
\[ \int_0^{\infty} e^{-r^2} r dr = \frac{1}{2} \int_0^{\infty} e^{-u} du = \frac{1}{2} \left[ -e^{-u} \right]_0^{\infty} = \frac{1}{2}(0 - (-1)) = \frac{1}{2} \] - Multiplying the results:
\[ I = \frac{\pi}{2} \cdot \frac{1}{2} = \frac{\pi}{4} \]

Step 4: Final Answer:
The value of the double integral is $\frac{\pi}{4}$.