Integral Formula

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

Can anyone prove that

odd result

(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

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:

$2^{2 r} \cos^{2 r} x \sin x = C (2 r, r) \sin x + \sum_{s = 1}^{r} C (2 r, x) (\sin (2 s + 1) x - \sin (2 s - 1) x)$ (*)

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

Now it is well-known that:

$\int_{0}^{\infty} (\sin m x) / x dx = π / 2$

for any positive $m$ . So:

$\int_{0}^{\infty} (\sin (2 s + 1) x - \sin (2 s - 1) x) / x dx = 0$

if $s \geq 1$ . So dividing (*) through by $x$ and integrating from 0 to $\infty$ we get:

$2^{2 r} \int_{0}^{\infty} (\cos^{2 r} x \sin x) / x dx = C (2 r, r) π / 2$

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

$\int_{0}^{\infty} (\cos^{2 r} x \sin x) / x dx = (- 1)^{r} C (- 1 / 2, r) π / 2$

Multiplying through by $z^{r}$ and summing from $r = 0$ to $\infty$ we get your result for $| z | < 1$ .