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?

  • 1
    $\begingroup$ 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
  • 3
    $\begingroup$ 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

\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*}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.