$2 Γ (n / 2 + 1) \int_{0}^{π / 2} \cos^{n} θ d θ = \sqrt{π} Γ ((n + 1) / 2)$

By Brad Rodgers on Monday, December 31, 2001 - 02:08 am:

How can I prove

\int_{0}^{π / 2} \cos^{n} θ d θ = \frac{Γ (\frac{n + 1}{2})}{n Γ (\frac{n}{2})} \sqrt{π}

I've tried a number of ways, the nearest attempt being contour integration around the quarter circle in the (positive,positive) quadrant, plus the lines connecting it to the origin (e.g.

[0, 1] + e^{i [0, π / 2]} + [i, 0]

) of the function

(z + z^{- 1})^{n} / (i 2^{n} z)

. For some reason, this didn't work; I got the answer of

(2 π / 2 n) C (n, n / 2)

for even

n

, and a similar result for odd. I will post my work if need be (if no one can come up with anything in a few days).
I'm really only concerned with proving this for integers, though a general result would be nice.

Thanks,

Brad

By Kerwin Hui on Monday, December 31, 2001 - 10:00 am:

Brad,

I will quickly sketch the proof. We are required to prove that $2 Γ (n / 2 + 1) \int_{0}^{π / 2} \cos^{n} θ d θ = \sqrt{π} Γ ((n + 1) / 2)$ , i.e.

$2 \int_{0}^{\infty} ξ^{n / 2} e^{- ξ} d ξ \int_{0}^{π / 2} \cos^{n} θ d θ = \sqrt{π} \int_{0}^{\infty} ψ^{(n - 1) / 2} e^{- ψ} d ψ$

Starting from the LHS, substitute $ξ = r^{2}$ and change variables to cartesian, do the $y$ -integral and use the substitution $ψ = x^{2}$ gives the RHS.

Kerwin

By Michael Doré on Monday, December 31, 2001 - 07:06 pm:

You can use a reduction formula method via integration by parts, taking u = cos^n-1 x and dv/dx = cos x, then use cos² x + sin² x = 1.