Consider the following infinite integral $$ \int_0^\infty q^{1-k} J_{2m+k} (q) J_\nu (qr) \, \mathrm{d} q \, , $$ where $\nu \in \{0,1\}$ and $m$ being a positive integer. Here $k$ and $r$ are both positive real number. I would like to make use of this integral for solving a physical situation involving dual integral equations.
My question is: under which conditions this integral vanishes in the case $r>1$.
It would be great if someone here could provide with some hints that could help a bit.
Thanks!
R
Take into account following equation: $$ \int^{\infty}_{0}t^{\mu-\nu+1}J_\mu (at)J_\nu(bt)dt=0,\quad \text{if }0<b<a,\\ \text{when }\Re(\nu)>\Re(\mu)>-1 \tag{1} $$ While equation of interest: $$ \int_0^\infty q^{1-k} J_{2m+k} (q) J_\nu (qr) \, \mathrm{d} q \tag{2} $$ As $m$ is positive integer and $k$ is positive real number, while $\nu\in\{0,1\}$, it is obvious that $\Re(2m+k)>\Re(\nu)$. Thus, for the sake of convenience let us rename variables: $$ q\to t,\quad \nu\to\mu,\quad 2m+k\to\nu,\quad 1-k\to1-\nu+2m $$
$$ \int_0^\infty t^{2m-\nu+1} J_\mu (tr) J_{\nu} (t) \, \mathrm{d} t \tag{3} $$ So, here we have $\Re(\nu)>\Re(\mu)>-1$ and $0<b<a$, where $b=1,a=r>1$. Only thing we need is to have hten is $2m=\mu$, but it is not the case, as $m$ is positive integer, while $\mu\in\{0,1\}$. Thus, to vanish according to the mentioned formula, initial integral should have different power of $q$.
It can be mentioned in comments $q^{1-k-2m+\nu}$, then integral vanish for all $\nu\in\{0,1\}$. Generally, it is $q^{1-k-2m+\epsilon}$, where $\epsilon\in\{0,1\}$ and integral vanish only if $\epsilon=\nu$, assumed that $\epsilon$ fixed and we vary $\nu$. (All definitions in this paragraph given in terms of initial variables).