0

Consider definite integral of sinc function to power $p$

$I(p) = \int_0^{\infty} = \left(\frac{\sin(x)}{x}\right)^p dx$

For simplest case $p=1$, well known result is $I(1) = \pi/2$.

How to derive result for $p=2$ and $p=3$?

Nigel1
  • 655
  • 1
    You can see the case for $p=2$ in https://19040886-a-62cb3a1a-s-sites.googlegroups.com/site/peeterjoot2/math2011/sincIntegral.pdf?attachauth=ANoY7coYkM22JTTrjq1loqpPPyOa6V4S8WtHUOA4C-w3MKB1ws34eHETSdwoaBlIjx1qQB1Rb3F2_ZA-ghaCHqR3fxp9A0g05zFXR_NEldRWeXS57tskp00wZcUYwOt8msduZkaDnGD1B8x19H5jVV8Rp4fQNQB4TL0dvtZJbN1fLyavbUhkjRwwQYBnW8Pi45eDH3rrpyXYe3GzJc_J9sJu-GEWNoVxism5FCQc8A-Xc5894ONvD4k%3D&attredirects=0. –  Dec 03 '16 at 14:09

2 Answers2

2

$I(p)$ has the following closed form: $$I(p)=\frac \pi {2^{p}}\left(p\sum_{k=0}^{\lfloor\frac{p-1}{2}\rfloor}\frac{(-2)^k(p-2k)^{p-1}}{k!(p-k)!}\right),~~\forall p\in\mathbb{N}^+.$$

See here or here.

Community
  • 1
Tianlalu
  • 5,177
1

For $n=2$

$$\int_0^\infty\frac{\sin^2 x}{x^2}dx=-\int_0^\infty\left(\frac{1}{x}\right)'\sin^2 xdx=-\left.\frac{1}{x}\sin^2 x\right|_{x=0}^{x=\infty}+\int_0^\infty \frac{2\sin x\cos x}{x}\\=\int_0^\infty \frac{\sin 2x}{x}=\frac{\pi}{2}$$

For $n=3$ the same trick: integrate by parts twice, then use $\cos^2 x=1-\sin^2 x$ and formula for $\sin(3x)=3\sin x-4\sin^3 x$ to reduce it to a linear combination of $\int_0^\infty\frac{\sin ax}{x}dx$

EDIT: To add some details on each integration by parts:

$$\int_0^\infty\frac{\sin^3 x}{x^3}dx=\frac{3}{2}\int_0^\infty\frac{\sin^2 x \cos x}{x^2}dx=\frac{3}{8}\int_0^\infty\frac{3\sin(3x)-\sin x}{x}dx=\frac{3\pi}{8}$$

Momo
  • 16,027