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I'm trying to solve an integral of the form: $$\int_0^{4\pi}dx\hspace{1mm}\mathrm{e}^{i(a\cos(x)+ib\sin(x)+cmx)}$$ where $a,b, c$ are constants, and $m\in\mathbb{N}$.

I've tried looking in the big book of integrals and series but couldn't find anything helpful. I thought maybe this integral has a solution in the form of a Bessel function. Individually, e.g.,

$$\int_0^{4\pi}dx\hspace{1mm}\mathrm{e}^{ia\cos(x)}$$ and $$\int_0^{4\pi}dx\hspace{1mm}\mathrm{e}^{-b\sin(x)+icmx}$$ have forms of Bessel functions and their variations. But together, there seems to be no solution to this.

I have also looked into changing variables to get rid of the cosine and sine terms, but still no luck.

Any suggestions?

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You integrals depend on Bessel function of the first kind: equation $(66)$ (the Jacobi-Anger expansion) is crucial. In particular, by setting $cm=C$, $\theta=\arctan\frac{B}{A}$ and $\rho=\sqrt{A^2+B^2}$,

$$\begin{eqnarray*} I(A,B,C) &=& \int_{0}^{4\pi}\exp\left(iA\cos(x)-B\sin(x)+iCx\right)\,dx\\ &=&2\int_{0}^{2\pi}\exp\left(i\sqrt{A^2+B^2} e^{i(x-\theta)}+iCx\right)\,dx\\&=&2e^{iC\left(\theta-\pi/2\right)}\int_{0}^{2\pi}\exp\left(\rho e^{ix}+iCx\right)\,dx \end{eqnarray*}$$ Now you may expand $\exp(\rho e^{ix})$ as a Taylor series in $e^{ix}$ and apply termwise integration (on $(0,2\pi)$) against $e^{iCx}$. If $C\in\mathbb{N}$, Parseval's theorem in the form $$ \int_{0}^{2\pi} e^{nix}e^{mix}\,dx = 2\pi\,\delta(m,n) $$ ensures that $I(A,B,C)$ has a nice and fast-converging series representation.

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  • $\begingroup$ I'm not sure how you got what you got, but I have something like this: $\int_0^{4\pi}dx\exp(i(\sqrt{A^2-B^2}\cosh(ix+y)+mcx))$ where $y=\text{arctanh}(B/A)$. I have tried integrating term by term, but this doesn't seem to help... I'm going to try and make the exponential contain sine or cosine terms then, as you said, use the Jacobi-Anger expansion. $\endgroup$ – Lewis Proctor Mar 1 '17 at 17:18
  • $\begingroup$ I have also reduced the integral to: $\int_0^{4\pi}\exp(i(\sqrt{A^2-B^2}\cos(x+y)+cmx))$. But because of the $x+y$ in the argument of the cosine, I can't use the integral representation of the Bessel function. $\endgroup$ – Lewis Proctor Mar 1 '17 at 18:43

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