| 2G(n/2+1)ò0p/2cosnqdq = |
Ö
|
p
|
G((n+1)/2)
|
By Brad Rodgers on Monday, December 31,
2001 - 02:08 am:
How can I prove
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] + ei[0,p/2] +[i,0]) of the
function (z+z-1 )n /(i2n z). For some reason, this didn't work; I got
the answer of (2p/2n )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
| 2G(n/2+1)ò0p/2cosnqdq = |
Ö
|
p
|
G((n+1)/2)
|
, i.e.
| 2ò0¥ xn/2 e-x dxò0p/2cosnqdq = |
Ö
|
p
|
ò0¥y(n-1)/2e-y dy
|
Starting from the LHS, substitute x = r2 and change variables to
cartesian, do the y-integral and use the substitution y = x2 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 = cosn-1 x and
dv/dx = cos x, then use cos2 x + sin2 x =
1.