Orthogonality of first and second kinds of solutions of Bessel's equation. Context
Bessel functions of the first kind, denoted as $J_\alpha(x)$, are solutions of Bessel's differential equation [1]. For appropriate boundary conditions the solutions satisfy an orthogonality relationship [1]. In particular:
$$
\int_0^1 x J_\alpha\left(x u_{\alpha,m}\right) J_\alpha\left(x u_{\alpha,n}\right) \,dx = \frac{\delta_{m,n}}{2} \left[J_{\alpha+1} \left(u_{\alpha,m}\right)\right]^2 .
$$
The Bessel functions of the second kind, denoted as $Y_\alpha(x)$, are solutions of the Bessel differential equation too.
Questions

*

*Is it true that, for appropriate boundary conditions the solutions satisfy an orthogonality relationship:
$$
\int_0^1 x Y_\alpha\left(x u_{\alpha,m}\right) Y_\alpha\left(x u_{\alpha,n}\right) \,dx = \frac{\delta_{m,n}}{2} \left[Y_{\alpha+1} \left(u_{\alpha,m}\right)\right]^2 \,\,?
$$


*For appropriate boundary conditions, do the solutions satisfy an orthogonality relationship:
$$
\int_0^1 x J_\alpha\left(x u_{\alpha,m}\right) Y_\alpha\left(x u_{\alpha,n}\right) \,dx \,\,?
$$
Bibliography
[1] Wikipedia contributors. "Bessel function." Wikipedia, The Free Encyclopedia. Wikipedia, The Free Encyclopedia, 16 Mar. 2021. Web. 17 Mar. 2021.
 A: 
For integral $\alpha$ and $\beta$, an orthogonality relation appears to be $$
\int_0^\infty J_\alpha(z)    \, N_\beta(z)    \frac {dz} z
= \frac 2 \pi \frac{\sin\left(\frac \pi 2 (\alpha-\beta) \right)}{\alpha^2 -\beta^2} \,\frac{\left[\cos{(\pi\beta)}
- 
 (-1)^\beta \right] }{\sin{(\pi\beta)}}
$$

I know that
$$\int_0^\infty J_\alpha(z) J_\beta(z) \frac {dz} z= \frac 2 \pi \frac{\sin\left(\frac \pi 2 (\alpha-\beta) \right)}{\alpha^2 -\beta^2}.
$$
I also know that for integral $\alpha$ and $\beta$, I can rewrite in terms of a Neuman fucntion as
$$\int_0^\infty J_\alpha(z)    \left[\frac{\sin{(\pi\,\beta)}\,N_\beta(z) + J_{-\beta}(z)  }{\cos{(\pi\beta)}}   \right] \frac {dz} z= \frac 2 \pi \frac{\sin\left(\frac \pi 2 (\alpha-\beta) \right)}{\alpha^2 -\beta^2}
$$
$$\int_0^\infty J_\alpha(z)    \left[\frac{\sin{(\pi\,\beta)}\,N_\beta(z) + J_{-\beta}(z)  }{\cos{(\pi\beta)}}   \right] \frac {dz} z= \frac 2 \pi \frac{\sin\left(\frac \pi 2 (\alpha-\beta) \right)}{\alpha^2 -\beta^2}
$$
$$\int_0^\infty J_\alpha(z)    \left[\frac{\sin{(\pi\,\beta)}\,N_\beta(z) + (-1)^\beta\,J_{\beta}(z)  }{\cos{(\pi\beta)}}   \right] \frac {dz} z= \frac 2 \pi \frac{\sin\left(\frac \pi 2 (\alpha-\beta) \right)}{\alpha^2 -\beta^2}
$$
Thus
$$
\int_0^\infty J_\alpha(z)    \left[\frac{\sin{(\pi\,\beta)}\,N_\beta(z) +  }{\cos{(\pi\beta)}}   \right] \frac {dz} z
+
\int_0^\infty J_\alpha(z)    \left[\frac{ (-1)^\beta\,J_{\beta}(z)  }{\cos{(\pi\beta)}}   \right] \frac {dz} z
= \frac 2 \pi \frac{\sin\left(\frac \pi 2 (\alpha-\beta) \right)}{\alpha^2 -\beta^2}
$$
Further,
Thus
$$
\int_0^\infty J_\alpha(z)    \left[\frac{\sin{(\pi\,\beta)}\,N_\beta(z) +  }{\cos{(\pi\beta)}}   \right] \frac {dz} z
+
\frac{(-1)^\beta}{\cos{(\pi\beta)}}\int_0^\infty J_\alpha(z)    \left[  J_{\beta}(z)    \right] \frac {dz} z
= \frac 2 \pi \frac{\sin\left(\frac \pi 2 (\alpha-\beta) \right)}{\alpha^2 -\beta^2}
$$
And
$$
\int_0^\infty J_\alpha(z)    \, N_\beta(z)    \frac {dz} z
= \frac 2 \pi \frac{\sin\left(\frac \pi 2 (\alpha-\beta) \right)}{\alpha^2 -\beta^2} \,\left[1 
- 
\frac{(-1)^\beta}{\cos{(\pi\beta)}}\right]\frac{\cos{(\pi\beta)}}{\sin{(\pi\beta)}}
$$
Thus, I find a orthogonality condition:
$$
\int_0^\infty J_\alpha(z)    \, N_\beta(z)    \frac {dz} z
= \frac 2 \pi \frac{\sin\left(\frac \pi 2 (\alpha-\beta) \right)}{\alpha^2 -\beta^2} \,\frac{\left[\cos{(\pi\beta)}
- 
 (-1)^\beta \right] }{\sin{(\pi\beta)}}
$$
