How do I show the above relation with Sturm-Liouville theory (assume the usual boundary conditions for the identity)? Here is what I have tried: if we start with
$$
\big(xJ\_a'(\ell x)
\big)'+\left(\ell^2-\frac{a^2}{x}\right)J\_a(\ell x)=0
$$
and
$$
\big(xJ\_a'(\ell' x)
\big)'+\left(\ell'^2-\frac{a^2}{x}\right)J\_a(\ell' x)=0
$$
and multiply the top expression by $xJ\_a(\ell' x)$ and bottom by $xJ\_a(\ell' x)$ and subtract the two equations,
$$
x\big(xJ\_a'(\ell x)
\big)'-x\big(xJ\_a'(\ell' x)
\big)'+x(\ell^2-\ell'^2)J\_a(\ell x)J\_a(\ell' x)=0.
$$
Then by rearrangement and integration over $[0, a]$ my answer is wrong. To be explicit, I get:
$$
\int\_0^axJ\_m(\ell x)J\_m(\ell'x)=\frac{a\big(J\_m'(\ell a)J\_m(\ell' a)-J\_m'(\ell' a)J\_m(\ell a)\big)}{\ell^2-\ell'^2}
$$
which is just barely wrong--- the correct result is
$$
\int\_0^axJ\_m(\ell x)J\_m(\ell'x)=\frac{a\big(\ell J\_m'(\ell a)J\_m(\ell' a)-\ell'J\_m'(\ell' a)J\_m(\ell a)\big)}{\ell^2-\ell'^2}
$$
which is zero (by the usual boundary conditions)


What am I doing wrong? Also-- is the reason we choose to multiply by $x\,\times$ the bessel function because of the "weight function" present in the Sturm–Liouville form of the Bessel function? Is this to ensure that our functions vanish at certain points?


Thank you!



Orthogonality of Bessel function $\int_0^bxJ_a(\ell x)J_a(\ell' x)=0$ for $\ell\neq\ell'$
For, $bl$ and $bl'$ being distinct zeroes of Bessel function , we have  

$(xJ\_a(lx)')'+(l^2x-q^2/X)J\_a(lx)=0$ and


$(xJ\_a(l'x)')'+(l'^2x-q^2/X)J\_a(l'x)=0$ .


Multiplying first with $J\_a(l'x)$ and second with $J\_a(l'x)$ and then subsratcting second from first ,


$-[J\_a(l'x)(xJ\_a(l'x))'-J\_a(lx)(xJ'\_a(l'x))']=x(l^2-l'^2)xJ\_a(lx)J\_a(l'x)$


Now, $J\_a(l'x)(xJ'\_a(l'x))'-J\_a(lx)(xJ'\_a(l'x))'=\frac{d}{dx}(xJ\_a(l'x)J'\_a(lx)-xJ\_a(lx)J'\_a(l'x)$


Hence, $\int\_{0}^{b} xJ\_a(lx)J\_a(l'x)=\frac{bJ\_a(l'b)J'\_a(lb)-bJ\_a(lb)J'\_a(l'b)}{l^2-l'^2}$.


