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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!

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Wouldn't it be easier to just say $a > 4$? –  Rahul Mar 8 '12 at 14:09
    
@RahulNarain Yes. –  Jonas Teuwen Mar 8 '12 at 14:23
3  
Downvoter... It is annoying. –  Ilya Mar 8 '12 at 18:10
    
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. –  user2468 Mar 9 '12 at 0:44
    
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. –  Jonas Teuwen Mar 9 '12 at 12:09

1 Answer 1

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

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1  
I will look into $a\not\in\mathbb{Z}$. –  robjohn Mar 8 '12 at 14:04
    
That would be nice :-). As I know this one ;-). I'm still a bit puzzled about your scaling though... –  Jonas Teuwen Mar 8 '12 at 17:18
    
I am puzzled about $\xi$ and $\zeta$. Is it a typo? –  Ilya Mar 8 '12 at 18:09
    
@Ilya If they would be the same, the integrals wouldn't be... I'm wondering. –  Jonas Teuwen Mar 8 '12 at 18:37
    
@Ilya: yes, I misread the $\zeta$ as a $\xi$. It is now fixed. Thanks. –  robjohn Mar 8 '12 at 18:53

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