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I have $\alpha_\max$ a real number between $0$ and $\frac\pi2$. Furthermore $\zeta$ and $\xi$ are positive real numbers.

Now I would like compute the integral

$$\int_0^{\alpha_\max} \mathrm{e}^{i \zeta \cos \alpha} \cos^n \alpha \, \sin^m \alpha J_k(\xi \sin \alpha) \, \mathrm{d}\alpha$$ For positive integers $n, m$ and $k$. Of course, these mess things up, so I first want to try a particular case where $k = 1$, $n = 1$ and $m = 2$. So I want to find the value of

$$\int_0^{\alpha_\max} \mathrm{e}^{i \zeta \cos \alpha} \cos \alpha \, \sin^2 \alpha J_1(\xi \sin \alpha) \, \mathrm{d}\alpha \tag{1}$$

Let's take the series expansion for our Bessel function $$J_1(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m+2)} {\left(\tfrac{1}{2}x\right)}^{2m+1}.$$

If we plug this in and pretend life is good we want to compute

$$\int_0^{\alpha_\max} \mathrm{e}^{i \zeta \cos \alpha} \, \sin^{2(m + 1) + 1} \alpha \, \cos \alpha \, \mathrm{d}\alpha$$

We also know $$\sin^{2(m + 1) + 1}\alpha = \frac{1}{4^{m + 1}} \sum_{k=0}^{m + 1} (-1)^{m - k + 1} \binom{2(m + 1) + 1}{k} \sin{((2(m - k + 1)\alpha)}$$

Alright! Let's just write this hideous integer in the $\sin$ as $m$ and we now want to compute

$$\int_0^{\alpha_\max} \mathrm{e}^{i \zeta \cos \alpha} \, \sin(m \alpha) \, \cos \alpha \, \mathrm{d}\alpha$$

This looks quite friendly, doesn't it? We can simplify it even more using a trigonometric identity which is well known to those who know it well. I'll recycle $m$ again. So we want to compute

$$\int_0^{\alpha_\max} \mathrm{e}^{i \zeta \cos \alpha} \, \sin(m \alpha) \, \mathrm{d}\alpha$$

Great! Messing a bit around with the integrand using Mathematica makes me suspect that the primitive is of the form

$$\sum_{k = 0}^ {m - 1} c_k \mathrm{e}^{i \zeta \cos \alpha} \cos(k \alpha)$$

So let's compute its derivative shall we?

$$\frac{\mathrm{d}}{\mathrm{d}\alpha}\sum_{k = 0}^ {m - 1} c_k \mathrm{e}^{i \zeta \cos \alpha} \cos(k \alpha) = \sum_{k = 0}^m [-k c_k - \frac{i \zeta}{2} c_{k - 1} + \frac{i \zeta}{2} c_{k + 1}] \sin(k \alpha) \mathrm{e}^{i \zeta \rho \cos \alpha}$$

So we can extract by equating the previous to the integrand: $$\begin{align} c_m &= 0;\\ -k c_k - \frac{i \zeta}{2} c_{k - 1} + \frac{i \zeta}{2} c_{k + 1} &= 0 \text{ for } 1 < k < m;\\ -\frac{i \zeta}{2} c_{m - 1} &= 1;\\ -c_1 + \frac{i \zeta}{2} c_2 - i \rho c_0 &= 0. \end{align}$$

I have not the slightest clue how to solve this. Making a generating function gives a horrendous differential equation which would even make the devil cry.

So, any suggestions on how to compute my dear integral $(1)$?

Edit: I believe that:

$$\newcommand{\d}{\, \mathrm{d}}\int J_1(\xi \sin\alpha) \d\alpha = \sum_{m = 0}^\infty \sum_{k = 0}^m \frac1{4^m} \frac{(-1)^m}{m!(m + 1)!} \frac{(-1)^{2m + 1 - k}}{2k + 1} {2m + 1 \choose m - k} \left(\frac{\xi}2 \right)^{2m + 1} \cos((2k + 1)\alpha).$$

Edit: It seems to be slightly off. I'll check again tomorrow it is past 2AM.

Edit: Now it is correct.

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With that $\alpha_{max}$ there, you are asking for the indefinite integral. Perhaps first do this case: $\int J_1(\sin\alpha)\,d\alpha$ –  GEdgar Mar 9 '12 at 16:16
    
@GEdgar Thanks. Do you mean that I should do this as a test case or will the result actually help me? –  Jonas Teuwen Mar 9 '12 at 16:43
    
@JonasTeuwen : \max is a standard operator name in $\TeX$. I changed the several occurrences of {\text{max}} to \max. When you use \max in a "displayed" as opposed to "inline" setting, then subscripts look like this: $\displaystyle\max_{x\in S}$. –  Michael Hardy Mar 9 '12 at 18:47
2  
@Michael Although it is not a solution to my problem it sure is useful! Thanks! :-) –  Jonas Teuwen Mar 9 '12 at 18:51
    
@GEdgar Well, done that! Now I need to find the relation between that and my problem... –  Jonas Teuwen Mar 10 '12 at 0:06
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