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Welcome to NRICH.

Integral Formula

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By Brad Rodgers on Monday, December 31, 2001 - 09:52 pm:

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

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By Michael Doré on Tuesday, January 01, 2002 - 06:14 pm:

Hi Brad,

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

2^2r cos^2rx sin x = C(2r,r) sin x + S^r _s=1C(2r,s)(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 ^¥(sin mx)/x dx = p/2

for any positive m. So:

ò₀ ^¥(sin(2s+1)x - sin(2s-1)x)/x dx = 0

if s³1. So dividing (*) through by x and integrating from 0 to ¥ we get:

2^2r ò₀ ^¥(cos^2rx sin x)/x dx = C(2r,r) p/2

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

ò₀ ^¥(cos^2rx sin x)/x dx = (-1)^rC(-1/2,r) p/2

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

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