What special function is this?

Question

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!

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. — , Mar 09 '12 at 00:44
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. — JT_NL, Mar 09 '12 at 12:09
I don't have sufficient time to look into this, but it seems to me that you'll need the Anger-Weber functions for these. The $\phi$ term makes things a tad inconvenient... — J. M. ain't a mathematician, Mar 19 '12 at 16:52

robjohn · Answer 1 · 2012-03-09T08:14:29.223

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 $n\in\mathbb{Z}$. $$ \begin{align} &\int_0^{2\pi}e^{i\zeta\cos(x+\phi)}\,\sin^2(x)\,\mathrm{d}x\\ &=\int_0^{2\pi}e^{i\zeta\cos(x)}\,\sin^2(x-\phi)\,\mathrm{d}x\\ &=\int_0^{2\pi}e^{i\zeta\cos(x)}\frac12(1-\cos(2(x-\phi)))\,\mathrm{d}x\\ &=\int_0^{2\pi}e^{i\zeta\cos(x)}\frac12(1-\cos(2x)\cos(2\phi)-\sin(2x)\sin(2\phi))\,\mathrm{d}x\\ &=\pi(J_0(\zeta)+\cos(2\phi)J_2(\zeta))\tag{4} \end{align} $$

edited Mar 09 '12 at 08:14

answered Mar 08 '12 at 13:44

robjohn

345,667

1

I will look into $a\not\in\mathbb{Z}$. – robjohn Mar 08 '12 at 14:04
That would be nice :-). As I know this one ;-). I'm still a bit puzzled about your scaling though... – JT_NL Mar 08 '12 at 17:18
I am puzzled about $\xi$ and $\zeta$. Is it a typo? – SBF Mar 08 '12 at 18:09
@Ilya If they would be the same, the integrals wouldn't be... I'm wondering. – JT_NL Mar 08 '12 at 18:37
@Ilya: yes, I misread the $\zeta$ as a $\xi$. It is now fixed. Thanks. – robjohn Mar 08 '12 at 18:53
@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). – robjohn Mar 08 '12 at 18:57
I think Mathematica is joking. The result is not independent on $\phi$, I'm sure. – JT_NL Mar 08 '12 at 19:32
@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. – SBF Mar 08 '12 at 21:26
@Ilya The derivative is not zero. The substitution is now fine, there was $\phi/a$. – JT_NL Mar 08 '12 at 21:35

What special function is this?

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