3

I needed a formula giving the value of the following integral; $$J(k)=\int_0^\pi dx\dfrac{\sin(x)^k}{x^k}$$ I didn't find any analytic expresion for it. I only found this interpolating function: $$J(k)=a\left(b^{\left(\dfrac{1}{k}\right)}\right)k^c-d$$ with: $a = 2.173,b = 0.8536,c = -0.4982,d = -0.002996$. valid until $k = 200$ with small error. Is t known any best fitting for $J(k)$? Thanks

Riccardo.Alestra
  • 10,746

1 Answers1

3

After using the trigonomy formulas for $(\sin x)^n$ and applying partial integration, I've got with $n\in\mathbb{N}$:

$$J(n)=\int\limits_0^\pi\left(\frac{\sin x}{x}\right)^n dx = \frac{1}{(n-1)!}\sum\limits_{k=0}^{\lfloor (n-1)/2\rfloor}(-1)^k{\binom n k}\left(\frac{n}{2}-k\right)^{n-1}\text{Si}((n-2k)\pi)$$

user90369
  • 11,696
  • 1
    Could you expand a little ? –  Oct 30 '19 at 10:59
  • Not explicitly, that's too much to write. I have used: $~\sin^n x = \frac{1}{2^n}{\binom n {n/2}} + \frac{1}{2^{n-1}}\sum\limits_{k=0}^{n/2-1}(-1)^{n/2-k}{\binom n k}\cos((n-2k)x)$ for even $n~$ and $~\sin^n x = \frac{1}{2^{n-1}}\sum\limits_{k=0}^{(n-1)/2}(-1)^{(n-1)/2}{\binom n k}\sin((n-2k)x)$ for odd $n~$ . Then dividing by $x^n$ and partial integration. – user90369 Oct 30 '19 at 11:07
  • No problem with the linearization of $\sin^k$, rather with the rest. –  Oct 30 '19 at 11:09
  • And I’ve used: $~\int\frac{\sin(ax)}{x^{2n-1}}dx = -\frac{\sin(ax)}{(2n-2)x^{2n-2}} + \frac{a}{2n-2}\int\frac{\cos(ax)}{x^{2n-2}}dx~$ and $~\int\frac{\cos(ax)}{x^{2n}}dx = -\frac{\cos(ax)}{(2n-1)x^{2n-1}}- \frac{a}{2n-1}\int\frac{\sin(ax)}{x^{2n-1}}dx~$ . – user90369 Oct 30 '19 at 11:13
  • The rest which has nothing to do with $Si(x)$ becomes $0$ by setting $x=0$ (unpleasant to calculate) and $x=\pi$ (simple to calculate). – user90369 Oct 30 '19 at 11:20