Question

The value of \(\int_0^\infty x^7 e^{-x^2} dx\) is

• A. 3
• B. 2
• C. 1
• D. 4

Hint:

For integrals of the form \(\int_{0}^{\infty} x^n e^{-x^2} dx\), use substitution \(t = x^2\) and the gamma function.
Remember \(\Gamma(n+1) = n!\) for integers.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a gamma function type integral.
Use substitution to transform it into a gamma function.

Step 2: Key Formula or Approach:
Let \(t = x^2\), then \(dx = \frac{dt}{2\sqrt{t}}\).
The integral becomes: \[ \int_{0}^{\infty} x^7 e^{-x^2} dx = \frac{1}{2} \int_{0}^{\infty} t^{3} e^{-t} dt. \]
Using the gamma function: \(\Gamma(n) = \int_{0}^{\infty} t^{n-1} e^{-t} dt\).

Step 3: Detailed Explanation:
Let \(t = x^2\), so \(x = t^{1/2}\), \(dx = \frac{1}{2} t^{-1/2} dt\).
Then \(x^7 = (t^{1/2})^7 = t^{7/2}\).
So \(x^7 dx = t^{7/2} \cdot \frac{1}{2} t^{-1/2} dt = \frac{1}{2} t^{3} dt\).
Thus, \[ I = \frac{1}{2} \int_{0}^{\infty} t^{3} e^{-t} dt = \frac{1}{2} \Gamma(4). \]
Since \(\Gamma(4) = 3! = 6\),
\(I = \frac{1}{2} \times 6 = 3\).
Thus, the value is 3, which is option (A).