Can anyone prove that



(or disprove it)?

This is a conjecture based off of numerical evidence only, so it could be false, but I've tested it pretty thoroughly, and I think it'll hold even for complex z. The only z I could evaluate it for analytically was z=0 (in which case the conjecture is true).

Brad

Hi Brad,

The identity is true for complex |z| < 1. First of all note that:

22rcos2rxsinx=C(2r,r)sinx+

r å s=1

C(2r,x)(sin(2s+1)x -sin(2s-1)x)

(*)

You can prove this by induction on r, or using complex exponentials.

Now it is well-known that:

ò₀^¥ (sinm x)/x dx=p/2

for any positive m. So:

ò₀^¥ (sin(2s+1)x-sin(2s-1)x)/x dx=0

if s ³ 1. So dividing (*) through by x and integrating from 0 to ¥ we get:

2^2rò₀^¥(cos^2r xsinx)/x dx=C(2r,r)p/2

Note that C(2r,r)=(-1)^r 2^2rC(-1/2,r), so we get:

ò₀^¥ (cos^2r xsinx)/x dx=(-1)^r C(-1/2,r)p/2

Multiplying through by z^r and summing from r=0 to ¥ we get your result for |z| < 1.