 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-1/16/c/solution

Write

B (n, n) = \int_{0}^{1} [x (1 - x)]^{n - 1} d x

and set

v = 2 x - 1

. Since

x (1 - x) = (1 - v^{2}) /4

and

d x = d v /2

B (n, n) = 2^{1 - 2 n} \int_{- 1}^{1} (1 - v^{2})^{n - 1} d v = 2^{2 - 2 n} \int_{0}^{1} (1 - v^{2})^{n - 1} d v .

In the last integral put

u = v^{2}

d v = d u / (2 u)

. This yields

B (n, n) = 2^{1 - 2 n} B (1/2, n)

, and part (a) interchanges the arguments. Hence

B (n, n) = \frac{B ( n , 1/2 )}{2 ^{2 n - 1}} .

The equal-argument beta integral identity holds for every positive real

n

 New to topics? Read the docs here!