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I'm trying to solve $\int_0^{2\pi} e^{i(a\cos\phi + b\sin\phi)} \cos\phi\ d\phi$ for a radiation problem in physics. In the special case that $b = 0$, this reduces to a Bessel function of the first kind, but I have no clue where to start for this. Any ideas?

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    $\begingroup$ Idea: $a \cos \phi+b \sin \phi = \sqrt{a^2+b^2} \cos(\phi-\phi_0)$ and then translate $\phi$. But you'll get a factor $\cos(\phi+\phi_0)$ outside which you would possibly have to develop. $\endgroup$ – H. H. Rugh Aug 1 '16 at 22:08
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    $\begingroup$ Bessel: $\displaystyle{\,\mathrm{J}_{1}}$. $\endgroup$ – Felix Marin Aug 1 '16 at 22:19
  • $\begingroup$ Split the exponential into two exponentials. Series expand the exponentials, most terms will vanish. $\endgroup$ – R. Rankin Aug 1 '16 at 23:25
  • $\begingroup$ @FelixMarin. Bessel $2 i \pi J_1(a)$ if $b=0$ I suppose. $b$ seems to be very problematic (at least to me). $\endgroup$ – Claude Leibovici Aug 2 '16 at 2:24
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\begin{align*} \int_0^{2\pi} e^{i(a\cos\phi+b\sin\phi)}\cos\phi \, d\phi &= \int_0^{2\pi} \left(-i \frac{\partial}{\partial a}\right) e^{i(a\cos\phi+b\sin\phi)}\,d\phi \\ &= -i \frac{\partial}{\partial a} \int_0^{2\pi} e^{i(a\cos\phi+b\sin\phi)}\,d\phi & \textrm{Leibniz rule} \\ &= -i \frac{\partial}{\partial a} \int_0^{2\pi} e^{i\sqrt{a^2+b^2}\cos(\phi-\phi_0)}\,d\phi & \textrm{see comment of H. H. Rugh} \\ &= -i \frac{\partial}{\partial a} \int_0^{2\pi} e^{i\sqrt{a^2+b^2}\cos t}\,d t & \textrm{periodicity of cosine} \\ &= -i \frac{\partial}{\partial a} 2\pi J_0\left(\sqrt{a^2+b^2}\right) & \textrm{standard integral} \\ &= \frac{2\pi i a J_1\left(\sqrt{a^2+b^2}\right)}{\sqrt{a^2+b^2}} & J_0'(x)=-J_1(x) \end{align*}

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