Thank you for your precious time!

Here is the integration I want to calculate:

$$\int_{0}^{2 \pi} \int_{0}^{2 \pi} e^{j 2 \pi\left[ \cos \theta- \cos \varphi+ \cos \left(\theta-\varphi\right)\right]} e^{j \theta} e^{-j \varphi} d \theta \, d \varphi$$

I think this integration is related to Bessel functions of the first kind, when we remove the term $\cos(\theta-\varphi)$ above:

$$\int_{0}^{2 \pi} \int_{0}^{2 \pi} e^{j 2 \pi\left[ \cos \theta- \cos \varphi\right]} e^{j \theta} e^{-j \varphi}\, d \theta d \varphi=\int_{0}^{2 \pi}e^{j (2 \pi \cos \theta+\theta)} d\theta \int_{0}^{2 \pi}e^{-j (2 \pi \cos \varphi+\varphi)}\, d\varphi$$ With the bessel integration: $$ J_{n}(x)=\dfrac{1}{2 \pi} \int_{-\pi}^{\pi} e^{j(n \tau-x \sin \tau)}\, d \tau $$ The integration would be easy to get.

But due to the $\cos(\theta-\varphi)$ term in the desired integration, I still can't find a good method.

Could you give me any advice?

Any help would be appreciated!


According to @Semiclasical's suggestion, I tried: $$e^{j c \cos \left(\theta-\varphi\right)}=\sum_{n=-\infty}^{\infty} j^{n} J_{n}\left(c\right) e^{i n\left(\theta-\varphi\right)}$$

So the integration would be:

$$\sum_{i=-\infty}^{\infty}J_n(c)j^n\int_{0}^{2 \pi} \int_{0}^{2 \pi} e^{j\left(a \cos \theta-b \cos \varphi\right)} e^{j\left(1+n\right) \theta} e^{-j\left(1+n\right) \varphi} d \theta d \varphi$$

And then it becomes: $$\sum_{i=-\infty}^{\infty} \dfrac{4 \pi^{2}}{j^{-n}} J_{n}\left(c\right) J_{1+n}\left(a\right) J_{-\left(1+n\right)}\left(b\right)$$

I think this is useful. But the integration becomes the summation of series. To be more specific, the integration becomes the summation of Bessel functions. It's still hard to get an analytical formula.

I tried to truncate this series, but the accuracy severely degraded. So for now, I think I should try something else. And currently, I'm planning to give Meijer-G function a try.

Could you give me more suggestions? Any help is appreciated!


I can't find useful Meijer-G functions for me.

But I find that the equation below is actually very accurate(even after truncation): $$\sum_{i=-\infty}^{\infty} \dfrac{4 \pi^{2}}{j^{-n}} J_{n}\left(c\right) J_{1+n}\left(a\right) J_{-\left(1+n\right)}\left(b\right)$$

So I decide to stick on the sum of bessel functions. Since it's hard to get the bessel function summation directly. I used the asymptotic form of bessel functions: $$ J_{\alpha}(z) \sim \frac{1}{\Gamma(\alpha+1)}\left(\frac{z}{2}\right)^{\alpha} $$

Then we can transform bessel series into power series. The summation might be easier. This should work only when $z$ is very small since this asymptotic form of bessel function only works when $z$ is very small.

Could you give me any advice?

Any help would be appreciated!

  • 1
    $\begingroup$ Along similar lines as you considered, the Jacobi-Anger expansion may come in handy. $\endgroup$ Mar 9, 2023 at 4:59
  • $\begingroup$ @Semiclassical Thanks! This gives me some inspiration. I will deliver an answer on this if possible. $\endgroup$ Mar 11, 2023 at 2:14
  • $\begingroup$ What is $j$? Is $j$ a complex unit, if so which one, a real constant, a complex constant, or something else entirely? $\endgroup$ Mar 11, 2023 at 11:16
  • 2
    $\begingroup$ @KevinDietrich Thanks for your reminding! For sure $j$ is a complex unit. Just like the $j$ in complex constant $a+bj$. $\endgroup$ Mar 11, 2023 at 11:21


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