Question

Let \[ J = \frac{1}{\pi} \int_0^1 t^{-\frac{1}{2}} (1 - t)^{\frac{3}{2}} \, dt. \] Then the value of \( J \) equals ..............

Hint:

Recognize integrals involving powers of \( t \) and \( 1 - t \) as Beta functions, which can be related to Gamma functions for easier evaluation.

Answer 1

Correct Answer: 0.35 - 0.4

Step 1: Recognize the form of the integral. 
The integral is in a form that resembles a Beta function. The Beta function is defined as: \[ B(x, y) = \int_0^1 t^{x-1} (1 - t)^{y-1} dt. \] By comparing the powers of \( t \) and \( 1 - t \), we see that this integral is a Beta function \( B\left( \frac{1}{2}, \frac{5}{2} \right) \).

Step 2: Apply the relationship between Beta and Gamma functions. 
Using the relation between Beta and Gamma functions: \[ B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)}, \] we can calculate the Beta function as: \[ B\left( \frac{1}{2}, \frac{5}{2} \right) = \frac{\Gamma\left( \frac{1}{2} \right) \Gamma\left( \frac{5}{2} \right)}{\Gamma\left( 3 \right)}. \] Using known values for the Gamma function \( \Gamma\left( \frac{1}{2} \right) = \sqrt{\pi} \) and \( \Gamma\left( \frac{5}{2} \right) = \frac{3}{4} \sqrt{\pi} \), we find: \[ B\left( \frac{1}{2}, \frac{5}{2} \right) = \frac{\sqrt{\pi} \cdot \frac{3}{4} \sqrt{\pi}}{2} = \frac{3\pi}{8}. \]

Step 3: Conclusion. 
Since the integral involves a factor of \( \frac{1}{\pi} \), we have: \[ J = \frac{1}{\pi} \cdot \frac{3\pi}{8} = 0.375 \]