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I just ran into integrals of the Bessel type, but which are unfortunately indefinite integrals, such as $$ f(t)=\int \cos(\gamma\cos(\omega t))\cos(\omega t)\mathrm dt. $$ I'm conscious of the fact that in a sense this is game over - if these integrals were doable in terms of getting elementary expressions for $f$, then Bessel functions would also be elementary and they are not. However, that's not a reason why there might not be standard ways to deal with a function like this one out there. Is $f(t)$ known in terms of other special functions, for example?

This is quite hard to google as most searches will just return indefinite integrals that have $J_\nu(x)$s in the integrand itself. If it's any help, this came up naturally while studying the motion of charged particles in oscillating electric and magnetic fields.

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  • $\begingroup$ In order to express such indefinite integrals, one would need “incomplete” Bessel functions, similar to the case of the beta and $\Gamma$ functions. Google returns about $1,000-1,600$ hits. $\endgroup$ – Lucian Oct 13 '15 at 9:33
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The best way I can see to deal with such an integral is to use the Fourier expansion

$$\cos{(\gamma \cos{\omega t})} = J_0(\gamma) + \sum_{n=1}^{\infty} (-1)^n J_{2 n}(\gamma) \cos{2 n \omega t} $$

Then

$$\int dt \, \cos{(\gamma \cos{\omega t})} \cos{\omega t} = \\J_0(\gamma) \frac{\sin{\omega t}}{\omega} + \frac1{2 \omega} \sum_{n=1}^{\infty} (-1)^n J_{2 n}(\gamma) \left [\frac{\sin{(2 n+1) \omega t}}{2 n+1} + \frac{\sin{(2 n-1) \omega t}}{2 n-1}\right ] + C$$

There might be a way to evaluate the sum in closed form.

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\begin{align}\int\cos(\gamma\cos\omega t)\cos\omega t \;\mathrm{d}t&=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\gamma^{2n}\cos^{2n+1}\omega t}{(2n)!}\;\mathrm{d}t\\ &=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\gamma^{2n}\cos^{2n}\omega t}{\omega(2n)!}\;\mathrm{d}(\sin\omega t)\\ &=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\gamma^{2n}(1-\sin^2\omega t)^n}{\omega(2n)!}\;\mathrm{d}(\sin\omega t)\\ &=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n\gamma^{2n}C_k^n(-1)^k\sin^{2k}\omega t}{\omega(2n)!}\; \mathrm{d}(\sin\omega t)\\ &=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}\gamma^{2n}n!\sin^{2k}\omega t}{\omega(2n)!k!(n-k)!}\;\mathrm{d}(\sin\omega t)\\ &=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{n+k}\gamma^{2n}n!\sin^{2k+1}\omega t}{\omega(2n)!k!(n-k)!(2k+1)}+\mathrm C\end{align}

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  • $\begingroup$ Oof, yeah, that's definitely doable. I was hoping for something a bit more packaged up. Thanks, though. (Note that MathJax also accepts \begin{align} constructs which look slightly better.) $\endgroup$ – E.P. Oct 16 '15 at 15:51

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