11
$\begingroup$

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.

  • 0
    @GEdgar Because the function might actually be easier to do? Consider $\int e^{-x^2}$ and $\int x e^{-x^2}$. Anyway, when I was sleeping I figured out that I forgot a factor.2012-03-10

0 Answers 0