I would like to know how to show the following: $$ \int_{0}^{2\pi}\int_{0}^{2\pi}H_{0}^{(1)}\left(\left|a+be^{i\theta}+ce^{i\phi}\right|\right)e^{im\theta}e^{in\phi}d\theta d\phi=4\pi^{2}(-1)^{m+n}e^{i\left(m\space arg\left(\frac{a}{b}\right)+n\space arg\left(\frac{a}{c}\right)\right)}J_{m}(|b|)J_{n}(|c|)H_{m+n}^{(1)}(|a|) $$ where $a$, $b$ and $c$ are any complex numbers satisfying $|b|+|c|<|a|$) and $m$ and $n$ are integers. The following Mathematica code suggests numerical agreement, with the ratio of the left hand side and the right hand side being very close to unity:
a = 2.33564 Exp[I \[Pi] 0.3215736];
b = 0.34268 Exp[I \[Pi] 2.5438987];
c = 1.21947 Exp[I \[Pi] 1.357315];
Table[{m, n,
NIntegrate[
HankelH1[0, Abs[a + b Exp[I \[Theta]] + c Exp[I \[Phi]]]] Exp[
I m \[Theta]] Exp[I n \[Phi]], {\[Theta], 0, 2 \[Pi]}, {\[Phi],
0, 2 \[Pi]}]/
N[4 \[Pi]^2 (-1)^(m + n) Exp[I (m Arg[a/b] + n Arg[a/c])] BesselJ[
m, Abs[b]] BesselJ[n, Abs[c]] HankelH1[m + n, Abs[a]]]}, {m, -2,
2}, {n, -2, 2}]
Thanks in advance for any help.
$\textbf{My attempt}$
Here I describe my progress thus far. I first use the following expansion for the Hankel function (convergent everywhere except $x=0$): $$ H_{0}^{(1)}(x)=\frac{2i}{\pi}\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k}}{2^{2k}(k!)^{2}}\left(\ln(x)+\frac{\pi}{2i}-\operatorname{ln(2)}+\gamma-H_{k}\right) $$ where $\gamma$ is the Euler-Mascheroni constant and $H_{k}$ is the $k^{th}$ Harmonic number. Thus we have the integral $$ \frac{2i}{\pi}\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2^{2k}(k!)^{2}}\left(\int_{0}^{2\pi}\int_{0}^{2\pi}\left|a+be^{i\theta}+ce^{i\phi}\right|^{2k}e^{im\theta}e^{in\phi}ln\left|a+be^{i\theta}+ce^{i\phi}\right|d\theta d\phi+\left(\frac{\pi}{2i}-\operatorname{ln(2)}+\gamma-H_{k}\right) \int_{0}^{2\pi}\int_{0}^{2\pi}\left|a+be^{i\theta}+ce^{i\phi}\right|^{2k}e^{im\theta}e^{in\phi}d\theta d\phi\right) $$ At this point, I write the integrand as a double contour integral around the unit circle (writing $z=e^{i\theta}$ and $\omega=e^{i\phi}$ and noting $z^{*}=\frac{1}{z}$ on the unit circle): $$ -\frac{2i}{\pi}\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2^{2k}(k!)^{2}}\left(\frac{1}{2}\int_{|z|=1}\int_{|\omega|=1}\left(a+bz+c\omega\right)^{k}\left(a+\frac{b}{z}+\frac{c}{\omega}\right)^{k}z^{m-1}\omega^{n-1}ln\left(a+bz+c\omega\right)dzd\omega+\frac{1}{2}\int_{|z|=1}\int_{|\omega|=1}\left(a+bz+c\omega\right)^{k}\left(a+\frac{b}{z}+\frac{c}{\omega}\right)^{k}z^{m-1}\omega^{n-1}ln\left(a+\frac{b}{z}+\frac{c}{\omega}\right)dzd\omega+\left(\frac{\pi}{2i}-\operatorname{ln(2)}+\gamma-H_{k}\right) \int_{|z|=1}\int_{|\omega|=1}\left(a+bz+c\omega\right)^{k}\left(a+\frac{b}{z}+\frac{c}{\omega}\right)^{k}z^{m-1}\omega^{n-1}dzd\omega\right) $$ The only contributions to the integrals are from the residues at the origins. For the middle integral I substitute $z\to\frac{1}{z}$ and $\omega\to\frac{1}{\omega}$ to avoid having to worry about the poles within the argument of the logarithm. Then we have $$ 8\pi i\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2^{2k}(k!)^{2}}\left(\frac{1}{2}coeff_{z^{-m}\omega^{-n}}\left(\left(a+bz+c\omega\right)^{k}\left(a+\frac{b}{z}+\frac{c}{\omega}\right)^{k}ln\left(a+bz+c\omega\right)\right)+\frac{1}{2}coeff_{z^{m}\omega^{n}}\left(\left(a+bz+c\omega\right)^{k}\left(a+\frac{b}{z}+\frac{c}{\omega}\right)^{k}ln\left(a+bz+c\omega\right)\right)+\left(\frac{\pi}{2i}-\operatorname{ln(2)}+\gamma-H_{k}\right) coeff_{z^{-m}\omega^{-n}}\left(\left(a+bz+c\omega\right)^{k}\left(a+\frac{b}{z}+\frac{c}{\omega}\right)^{k}\right)\right) $$ At this stage, I could work out the coefficients exactly by expanding out the expressions, but I then end up with huge sums of products of many factorials, so I don't really want to do this. Instead, after substituting back $z\to\frac{1}{z}$ and $\omega\to\frac{1}{\omega}$ in the middle term and regrouping terms, we have $$ 8\pi i\space coeff_{z^{-m}\omega^{-n}}\left(\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2^{2k}(k!)^{2}}\left(a+bz+c\omega\right)^{k}\left(a+\frac{b}{z}+\frac{c}{\omega}\right)^{k}\left(\frac{1}{2}ln\left(a+bz+c\omega\right)+\frac{1}{2}ln\left(a+\frac{b}{z}+\frac{c}{\omega}\right)+\left(\frac{\pi}{2i}-\operatorname{ln(2)}+\gamma-H_{k}\right) \right)\right) $$ The sum can be evaluated exactly through $\sum_{k=0}^{\infty}\frac{(-1)^{k}y^{k}}{(k!)^{2}}=J_{0}(2\sqrt{y})$ (the Bessel function of the first kind), so $$ 8\pi i\space coeff_{z^{-m}\omega^{-n}}\left(J_{0}\left(\sqrt{\left(a+bz+c\omega\right)\left(a+\frac{b}{z}+\frac{c}{\omega}\right)}\right)\left(\frac{1}{2}ln\left(a+bz+c\omega\right)+\frac{1}{2}ln\left(a+\frac{b}{z}+\frac{c}{\omega}\right)+\left(\frac{\pi}{2i}-\operatorname{ln(2)}+\gamma-H_{k}\right) \right)\right) $$ At this point I'm a bit stuck.