4
$\begingroup$

Assume that $\zeta$ is a positive real number and $a = \frac{2 \pi}{\alpha_{\text{max}}}$ for $0 < \alpha_{\text{max}} < \frac{\pi}{2}$. In other words $a > 4$.

Is there a special function that when evaluated in a certain point is equal to

$$\int_0^{2 \pi} \textrm{e}^{i \zeta \cos(ax + \phi)} \, \sin^2(ax) \, \mathrm{d}x?$$

If $a$ would be a nice and an integer life would be good. Now I don't know!

  • 0
    Wouldn't it be easier to just say $a > 4$?2012-03-08
  • 0
    @RahulNarain Yes.2012-03-08
  • 3
    Downvoter... It is annoying.2012-03-08
  • 0
    The title "What special function is this?" does not tell much about the content of the post. Titles appear on the frontpage, Related sidebar, and on Google search results. The readers of such three places need a good, differential, idea on the content of the post.2012-03-09
  • 0
    I don't agree. If you don't know anything about special functions the addition will help nothing. Google search does not parse the LaTeX anyway.2012-03-09
  • 0
    I don't have sufficient time to look into this, but it seems to me that you'll need the [Anger-Weber functions](http://dlmf.nist.gov/11.10) for these. The $\phi$ term makes things a tad inconvenient...2012-03-19

1 Answers 1

4

For $a\in\mathbb{Z}$, we have $$ \int_0^{2\pi}e^{i\zeta\cos(ax+\phi)}\,\sin^2(ax)\,\mathrm{d}x=\int_0^{2\pi}e^{i\zeta\cos(x+\phi)}\,\sin^2(x)\,\mathrm{d}x\tag{1} $$ Mathematica says that this is $$ \frac{2\pi}{\zeta}\operatorname{BesselJ}(1,\zeta)=\frac{2\pi}{\zeta}\operatorname{J}_1(\zeta)\tag{2} $$ for $\zeta\in\mathbb{R}^+$.


It appears that Mathematica is not correct; i.e. The integral in $(1)$ is not independent of $\phi$ as $(2)$ would indicate. For now, we will keep the same assumptions ($a\in\mathbb{Z}$ and $\zeta\in\mathbb{R}^+$). We will also use $$ \begin{align} \int_0^{2\pi}e^{i\zeta\cos(x)}\cos(nx)\,\mathrm{d}x&=2\pi\,i^nJ_n(\zeta)\\ \int_0^{2\pi}e^{i\zeta\cos(x)}\sin(nx)\,\mathrm{d}x&=0 \end{align}\tag{3} $$ for $n\in\mathbb{Z}$. $$ \begin{align} &\int_0^{2\pi}e^{i\zeta\cos(x+\phi)}\,\sin^2(x)\,\mathrm{d}x\\ &=\int_0^{2\pi}e^{i\zeta\cos(x)}\,\sin^2(x-\phi)\,\mathrm{d}x\\ &=\int_0^{2\pi}e^{i\zeta\cos(x)}\frac12(1-\cos(2(x-\phi)))\,\mathrm{d}x\\ &=\int_0^{2\pi}e^{i\zeta\cos(x)}\frac12(1-\cos(2x)\cos(2\phi)-\sin(2x)\sin(2\phi))\,\mathrm{d}x\\ &=\pi(J_0(\zeta)+\cos(2\phi)J_2(\zeta))\tag{4} \end{align} $$

  • 1
    I will look into $a\not\in\mathbb{Z}$.2012-03-08
  • 0
    That would be nice :-). As I know this one ;-). I'm still a bit puzzled about your scaling though...2012-03-08
  • 0
    I am puzzled about $\xi$ and $\zeta$. Is it a typo?2012-03-08
  • 0
    @Ilya If they would be the same, the integrals wouldn't be... I'm wondering.2012-03-08
  • 0
    @Ilya: yes, I misread the $\zeta$ as a $\xi$. It is now fixed. Thanks.2012-03-08
  • 0
    @Jonas: re scaling: If $a\in\mathbb{Z}$, then substituting $x\to x/a$ changes the range of integration to $[0,2\pi a]$ and that is offset by the $\mathrm{d}x/a$ (multiply by $a$ and divide by $a$, if I am not mistaken).2012-03-08
  • 0
    I think Mathematica is joking. The result is not independent on $\phi$, I'm sure.2012-03-08
  • 0
    @Jonas The result would be independent only if the derivative is zero everywhere. Have you tried Maple? The substitution by Rob looks perfectly fine for me.2012-03-08
  • 0
    @Ilya The derivative is not zero. The substitution is *now* fine, there was $\phi/a$.2012-03-08