Orthogonality of different Bessel functions According to Orthogonality condition, it is true that:
$\int_0^b   xJ_0(\lambda_nx)J_0(\lambda_mx)dx = 0$   if   $m \not=n $
What is the result for :
$\int_0^b   xJ_0(\lambda_nx)Y_0(\lambda_mx)dx = ?$   
For two cases:
if   $m \not=n $
if $m=n$
 A: For any $\mu, \nu > 0$, let $\tilde{J}(z) = J_0(\mu z)$ and $\tilde{Y}(z) = Y_0(\nu z)$. They satisfies the ODE:
$$
\frac{d}{dz}\big[z\tilde{J}'(z)\big] + \mu^2 z \tilde{J}(z) = 0
\quad\text{ and }\quad
\frac{d}{dz}\big[z\tilde{Y}'(z)\big] + \nu^2 z \tilde{Y}(z) = 0
$$
Mutiply $1^{st}$ ODE by $\tilde{Y}$, the $2^{nd}$ by $\tilde{J}$ and subtract them, we get:
$$\frac{d}{dz}\big[z\tilde{J}'(z)\big]\tilde{Y}(z) 
- \frac{d}{dz}\big[z\tilde{Y}'(z)\big]\tilde{J}(z) + (\mu^2-\nu^2) z\tilde{J}(z)\tilde{Y}(z) = 0
$$
This implies
$$\begin{align}
z \tilde{J}(z)\tilde{Y}(z) 
= & \frac{1}{\nu^2-\mu^2}\frac{d}{dz}\left[z\left( \tilde{J}'(z)\tilde{Y}(z) - \tilde{Y}'(z)\tilde{J}(z)\right)\right]\\
= & \frac{1}{\mu^2-\nu^2}\frac{d}{dz}\Big[\mu z J_1(\mu z) Y_0(\nu z) - \nu z Y_1(\nu z)J_0(\mu z)\Big]
\end{align}
$$
Notice 
$$\lim_{z\to 0} z Y_0(z) = 0, \quad \lim_{z\to 0} z Y_1(z) = -\frac{2}{\pi}$$
and $J_0(z), J_1(z)$ are regular at $z = 0$, we get:
$$\begin{align} &\int_{0}^{b} z J_0(\mu z)Y_0(\nu z) dz\\
= & \frac{1}{\mu^2-\nu^2}\lim_{\epsilon\to 0}\Big[
\mu z J_1(\mu z) Y_0(\nu z) - \nu z Y_1(\nu z)J_0(\mu z) \Big]_{\epsilon}^b\\
= & \frac{1}{\nu^2-\mu^2}\left[
\frac{2}{\pi} - b \left( \mu J_1(\mu b) Y_0(\nu b) - \nu Y_1(\nu b)J_0(\mu b) \right)\tag{*1}
\right]\end{align}
$$
If $\lambda_m$ and $\lambda_n$ are distinct roots of $J_0(\lambda b)$, this reduces to:
$$\color{firebrick}{\int_{0}^{b} z J_0(\lambda_n z)Y_0(\lambda_m z) dz 
= \frac{1}{\lambda_m^2 - \lambda_n^2}\left[\frac{2}{\pi} - \lambda_n b J_1(\lambda_n b)Y_0(\lambda_m b)\right]}
$$
For the remaining $\lambda_m = \lambda_n$ case, we need to use the fact:
$$J_1(z) Y_0(z) - Y_1(z) J_0(z) = Y_0'(z)J_0(z) - J_0'(z)Y_0(z)$$
is a Wronskian for the pair of solutions
$J_0(z)$, $Y_0(z)$ of the $k = 0$-th Bessel ODE:
$$\frac{d^2\phi(z)}{dz^2} + \frac{1}{z}\frac{d\phi(z)}{dz}  + ( 1 - \frac{k^2}{z^2} ) \phi(z) = 0$$
General theory of $2^{nd}$ order ODE tell us it is proportional to $\frac{1}{z}$.
Evaluate it at $z \sim 0$ shows that it is equal to $\frac{2}{\pi z}$. This means in $(*1)$,
we can take the limit $\mu \to \nu$ to obtain:
$$\begin{align}\int_{0}^{b} z J_0(\nu z)Y_0(\nu z) dz 
= & \frac{b}{2\nu}
\left( 
\frac{d}{d\mu} \big[\mu J_1(\mu b)\big]_{\mu\to\nu} Y_0(\nu b) - 
\nu Y_1(\nu b)\frac{d}{d\mu}\big[J_0(\mu b)\big]_{\mu\to\nu}
\right)\\
= & \frac{b^2}{2}\big( J_0(\nu b)Y_0(\nu b) + J_1(\nu b)Y_1(\nu b)\big)
\end{align}
$$
When $\lambda_m$ is a root of $J_0(\lambda b)$, this simplifies to:
$$\color{firebrick}{\int_{0}^{b} z J_0(\lambda_m z)Y_0(\lambda_m z) dz  = \frac{b^2}{2} J_1(\lambda_m b)Y_1(\lambda_m b)}$$
A: Presumably $\lambda_j$ are the zeros of $J_0(\lambda b)$.  The answer is not $0$.
For example, with $b=1$,  Maple says
$$\int_0^1 x J_0(\lambda_1 x) Y_0(\lambda_2 x)\ dx \approx 0.042926296743804723657$$
I doubt that there is a closed-form formula.  
