Question

The value of the integral \( \int_0^\infty \int_0^x x e^{-x^2/y} dy \, dx \) is:

• A. 1
• B. \( \frac{3}{2} \)
• C. 0
• D. \( \frac{1}{2} \)

Hint:

When a double integral seems intractable, always consider changing the order of integration. This is the most common trick for difficult double integrals in exams. Be sure to correctly determine the new limits by sketching or analyzing the inequalities defining the region.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This is an improper double integral. The integrand \( e^{-x^2/y} \) is difficult to integrate with respect to \(y\). This suggests that changing the order of integration might simplify the problem.

Step 2: Key Formula or Approach:
The region of integration is given by \( 0 \le y \le x \) and \( 0 \le x<\infty \). This describes the infinite region in the first quadrant between the lines \(y=0\) and \(y=x\). To change the order of integration, we fix \(y\) first and let \(x\) vary. - The minimum \(y\) value is 0. The maximum is \(\infty\). - For a fixed \(y\), \(x\) goes from the line \(x=y\) to \(x=\infty\). So the new limits are \( y \le x<\infty \) and \( 0 \le y<\infty \). The integral becomes: \[ \int_0^\infty \int_y^\infty x e^{-x^2/y} dx \, dy \]
Step 3: Detailed Explanation:
Let's evaluate the inner integral with respect to \(x\): \[ I_x = \int_y^\infty x e^{-x^2/y} dx \] Use a u-substitution. Let \( u = -x^2/y \). Then \( du = (-2x/y) dx \), which gives \( x dx = -\frac{y}{2} du \). Change the limits of integration for \(u\): - When \( x=y \), \( u = -y^2/y = -y \). - When \( x \to \infty \), \( u \to -\infty \). The inner integral becomes: \[ I_x = \int_{-y}^{-\infty} e^u \left(-\frac{y}{2} du\right) = \frac{y}{2} \int_{-\infty}^{-y} e^u du \] \[ = \frac{y}{2} [e^u]_{-\infty}^{-y} = \frac{y}{2} (e^{-y} - \lim_{a \to -\infty} e^a) = \frac{y}{2}(e^{-y} - 0) = \frac{y}{2}e^{-y} \] Now, substitute this result back into the outer integral: \[ \int_0^\infty \left( \frac{y}{2}e^{-y} \right) dy = \frac{1}{2} \int_0^\infty y e^{-y} dy \] This is the Gamma function \( \Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt \). Here, \(z-1=1\), so \(z=2\). The integral is \( \frac{1}{2} \Gamma(2) \). Since \( \Gamma(n)=(n-1)! \) for integers \(n\), \( \Gamma(2) = 1! = 1 \). So the value is \( \frac{1}{2} (1) = \frac{1}{2} \). Alternatively, we can use integration by parts for \( \int y e^{-y} dy \): Let \(u=y, dv=e^{-y}dy\). Then \(du=dy, v=-e^{-y}\). \[ \int y e^{-y} dy = -ye^{-y} - \int (-e^{-y})dy = -ye^{-y} - e^{-y} \] \[ \int_0^\infty y e^{-y} dy = [-e^{-y}(y+1)]_0^\infty = \lim_{b \to \infty} [-e^{-b}(b+1)] - (-e^0(0+1)) = 0 - (-1) = 1 \] So the final answer is \( \frac{1}{2}(1) = \frac{1}{2} \).

Step 4: Final Answer:
The value of the integral is \( \frac{1}{2} \).