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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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    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

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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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    @Ilya The derivative is not zero. The substitution is *now* fine, there was $\phi/a$.2012-03-08