A transformation of Bessel Y Zero $y_{v,x}$ inverts $\arg H^{(1)}_v(x)$ with the argument function and the first Hankel function. Also, another one inverts $\frac{Y_v(x)}{J_v(x)}$ with Bessel J and Bessel Y. I am inexperienced with Fourier series, but it gives a potentially interesting series of $y_{v,x+1}$. We use the ratio of Bessel functions for the first integral with, for simplicity, $|x|\le \frac L2,|L|<1$:
$$y_{v,x+1}=\frac1L \sum_{n=-\infty}^\infty \int_{-\frac L2}^\frac L2 y_{v,x+1}e^{-\frac{2\pi in x}L}dxe^\frac{2\pi i nx}L$$
Now substitute $x\to y_{v,x+1}$: $$\int_{-\frac L2}^\frac L2 y_{v,x+1}e^{-\frac{2\pi in x}L}dx=\frac1\pi\int_{y_{v,1-\frac L2}}^{y_{v,\frac L2+1}}x e^{-\frac{2in \tan^{-1}\left(\frac{Y_v(x)}{J_v(x)}\right)}L} d\left(\tan^{-1}\left(\frac{Y_v(x)}{J_v(x)}\right)\right)$$
We simplify and use the polar form of complex numbers:
$$\frac1\pi\int_{y_{v,1-\frac L2}}^{y_{v,\frac L2+1}}x e^{-\frac{2in \tan^{-1}\left(\frac{Y_v(x)}{J_v(x)}\right)}L} d\left(\tan^{-1}\left(\frac{Y_v(x)}{J_v(x)}\right)\right) = \frac2{\pi^2}\int_{y_{v,1-\frac L2}}^{y_{v,\frac L2+1}}\frac{\left(\frac{2i}{i-\frac{Y_v(x)}{J_v(x)}}-1\right)^\frac nL}{J_v^2(x)+Y_v^2(x)}dx$$
Here is $\arg\left(H_v^{(1)}(x)\right)$ representation with the sign, second Hankel, and Bessel K functions, and the substitution $x\to y_{v,x+1}$:
$$\begin{align}\int_{-\frac L2}^\frac L2 y_{v,x+1}e^{-\frac{2\pi in x}L}dx= \frac1\pi\int_{y_{v,1-\frac L2}}^{y_{v,\frac L2+1}} xe^{-\frac{2 in \arg\left(H_v^{(1)}(x)\right)}L}d\left(\arg\left(H_v^{(1)}(x)\right)\right)= \frac2{\pi^2}\int_{y_{v,1-\frac L2}}^{y_{v,\frac L2+1}} \frac{dx}{\text{sgn}^\frac{2n}L\left(H_v^{(1)}(x)\right) |H_v^{(1)}(x)|^2}dx=\frac2{\pi^2}\int_{y_{v,1-\frac L2}}^{y_{v,\frac L2+1}}\frac{| H_v^{(1)}(x)|^{\frac{2n}L-2}}{H_v^{(1)}(x)^\frac{2n}L}dx=\frac2{\pi^2}\int_{y_{v,1-\frac L2}}^{y_{v,\frac L2+1}}\frac{H^{(2)}_v(x)^{\frac nL-1}}{H_v^{(1)}(x)^{\frac nL+1}}dx=\frac i2(-1)^{v\left(\frac nL+1\right)+1}\int_{iy_{v,1-\frac L2}}^{iy_{v,\frac L2+1}}\frac{K_v(x)^{\frac nL-1}}{K_v(-x)^{\frac nL+1}}dx\end{align}$$
These coefficients are integrals over zeros of the imaginary part of powers and ratios of Bessel functions:
How can you evaluate $\begin {align}\displaystyle \frac i2(-1)^{v\left(\frac nL+1\right)+1}\int_{iy_{v,1-\frac L2}}^{iy_{v,\frac L2+1}}\frac{K_v(x)^{\frac nL-1}}{K_v(-x)^{\frac nL+1}}dx= \frac2{\pi^2}\int_{y_{v,1-\frac L2}}^{y_{v,\frac L2+1}}\frac{H^{(2)}_v(x)^{\frac nL-1}}{H_v^{(1)}(x)^{\frac nL+1}}dx =\frac2{\pi^2}\int_{y_{v,1-\frac L2}}^{y_{v,\frac L2+1}}\frac{\left(\frac{2i}{i-\frac{Y_v(x)}{J_v(x)}}-1\right)^\frac nL}{J_v^2(x)+Y_v^2(x)}dx\end{align}$ so we have an explicit series expansion of the Bessel Y zero function? If the complex Fourier series should not be used, then we take the imaginary and real parts of the integrals and use other Fourier series formulas:
- A series expansion, or any form besides an integral, of any one of the integrals presented in this question will suffice.
- If it helps, making the integrals indefinite and evaluating any one presented in this question is ok. Please do not approximate.