Recurrence for $\int \left(\frac{\sin x}{x}\right)^n \, \mathrm{d}x$

Asked Nov 04 '15 at 07:59

Active Nov 04 '15 at 09:27

Viewed 214 times

I was playing around and wanted to consider the integral of $$I_n = \int_0^{\infty} \left(\frac{\sin x}{x}\right)^n \, \mathrm{d}x$$ using parts with $u = \sin^n x \implies \mathrm{d}u = n\cos x \sin^{n-1}x$ and $\mathrm{d}v = x^{-n} \implies v = \frac{x^{1-n}}{1-n}$

This gives $$I_n = \frac{n}{n-1}\int_0^{\infty} \left(\frac{\sin x}{x}\right)^{n-1} \cos x \, \mathrm{d}x$$

Which looks so close to $I_{n-1}$ but for that annoying $\cos x$ there, any help?

Note: I wanted to go down this route because I think I can note a pattern, as such: $$I_1 = \frac{\pi}{2} \\ I_2 = \frac{\pi}{2} \\ I_3 = \frac{3\pi}{8} \\ I_4 = \frac{\pi}{3} \\ I_5 = \frac{115\pi}{384} \\ I_6 = \frac{11\pi}{40}$$

edited Nov 04 '15 at 08:04

Eugene Zhang

16,805

asked Nov 04 '15 at 07:59

Zain Patel

16,802

3

try to use the integration by parts once again – Michael Medvinsky Nov 04 '15 at 08:16
@MichaelMedvinsky With $u = \sin^{n-1}x \cos x$ and $\mathrm{d}v = x^{1-n}$? – Zain Patel Nov 04 '15 at 08:17
no, leave the cosinus for du to get $\cos^2x=1-\sin^2x$ – Michael Medvinsky Nov 04 '15 at 08:22
@MichaelMedvinsky so $u = \sin^{n-1} x$ and $\mathrm{d}v = \frac{\cos x}{x^{n-1}}$? – Zain Patel Nov 04 '15 at 08:24
2

another integration by parts didn't work for me yet, also expected it would – Michael Medvinsky Nov 04 '15 at 09:32
It seems that one has to deal with the more general integral
$$I_{m, n} = \int_0^\infty \frac{\sin^m x}{x^n} dx,\ \ \ \text{where } m\ge n.$$

By using integration by part twice, one obtains

$$I_{m, n}= \frac{m(m-1)}{(n-1)(n-2)} I_{m-2, n-2} - \frac{m^2}{(n-1)(n-2)} I_{m,n-2}.$$
– Nov 04 '15 at 09:38
But I am still working on the base case: $I_{m, 1}$ and $I_{m,2}$. – Nov 04 '15 at 09:40
1

It might be a good idea (I did not test it completely) to use Fourier transforms and convolutions. – mickep Nov 04 '15 at 09:43
@JohnMa what advantage does working out the more general integral bring? :) – Zain Patel Nov 04 '15 at 09:43
@mickep - I know what those are, but they are a tad too advanced a technique for me to use. ;-) – Zain Patel Nov 04 '15 at 09:43
@ZainPatel Because after integration by part twice one has that term $I_{m,n-2}$ unfortunately. – Nov 04 '15 at 09:45
@JohnMa - ah, okay, yeah, that makes sense! – Zain Patel Nov 04 '15 at 09:47
@MichaelMedvinsky. I deleted my answer. Thanks for the link. But the summation disappears because using Eulerian numbers which are given explicitly bya sum ! Cheers – Claude Leibovici Nov 04 '15 at 10:20
@Claude Leibovici sure! I will just put the link here so it available http://mathworld.wolfram.com/SincFunction.html – Michael Medvinsky Nov 04 '15 at 10:30
See here: http://math.stackexchange.com/questions/146741/using-contour-integration-or-other-means-is-there-a-way-to-find-a-general-form – Cody Nov 04 '15 at 12:52

Recurrence for $\int \left(\frac{\sin x}{x}\right)^n \, \mathrm{d}x$

0 Answers0