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We know that (from e.g. here) $$ \int_0^{2\pi}{\rm e}^{{\rm i}\left[a\sin\left(x\right) + b\cos\left(x\right)\right]}{\rm d}x = 2\pi\operatorname{J}_{0}\left(\sqrt{a^{2} + b^{2}}\right) $$ where $\operatorname{J}_0$ is the Bessel function.

I'm trying to see if we can express $\displaystyle\int_0^{\pi}{\rm e}^{{\rm i}\left[a\sin\left(x\right) + b\cos\left(x\right)\right]}{\rm d}x$ in a similar fashion.

I searched for different sources, but I couldn't find one that does not involve infinite summations over Bessel functions. Any help would be very much appreciated !.

One possible idea: perhaps first expanding exponentials using real-valued Jacobi-Anger expansion, which gives integrals of, e.g. product of $\sin (2mx)$ and $\cos(2nx)$, which forces $m = n$. This will make the integral to involve terms something like $\sum J_{2n}(a) J_{2n}(B)$ that can be represented in terms of $J_0(\cdot)$ using Neumann's addition formula. However, I'm not sure if this is a right approach..

userflux9674
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    By symmetry if $a = 0$ the value of the integral is $\pi J_0(b)$. If $b = 0$, then the value is $\pi J_0(a) + \pi i H_0(a)$, where $H_0$ is the Struve function. https://en.wikipedia.org/wiki/Struve_function I don't immediately see a way to evaluate the integral for arbitrary $(a, b)$ without series. – Travis Willse May 06 '24 at 01:36
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    $\int_0^{2\pi}e^{i(a\cos(x)+b\sin(x))}dx= 2\text{Re}\int_0^\pi e^{i(a\cos(x)+b\sin(x))}dx$. You can try using the harmonic addition formula too – Тyma Gaidash May 06 '24 at 01:39
  • @TravisWillse Thank you, interesting; but that doesn't imply that the integral should be $\pi J_0(b) + \pi J_0(a) + \pi i H_0(a)$, right? – userflux9674 May 06 '24 at 04:55
  • @ТymaGaidash Great point; is there any way to get the imaginary part of the integral as well? – userflux9674 May 06 '24 at 04:55
  • @userflux9674 No, I don't think so. – Travis Willse May 06 '24 at 17:07
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    After some explicit calculations, it seems like the integral is $\pi J_0(\sqrt{a^2+b^2}) + i\pi J_0(b) H_0(a) $ (might be missing some constant factors). Both of your comments helped me a lot. – userflux9674 May 06 '24 at 19:13

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