How to prove this equation with Bessel function and Laguerre function? I find the following equation in a literature. I need to prove it, but I find it is very difficult of me. Any one can help me to prove this equation? Your help or comments are highly appreciated. 
$\int_0^\infty  {\mathop{\rm e}\nolimits}  xp[ - ax^2 ]J_\nu  (2bx)x^{2n + \nu  + 1} dx = \frac{{n!}}{2}b^\nu  a^{ - n - \nu {\rm{ - }}1} \exp ( - \frac{{b^2 }}{a})L_n^v (\frac{{b^2 }}{a})$
Where  $J_v (2bx) \equiv \sum\limits_{{\rm{k = 0}}}^\infty  {\frac{{( - 1)^k }}{{k!\Gamma (k + v + 1)}}} (bx)^{2k + v} {\rm{ = }}\sum\limits_{{\rm{k = 0}}}^\infty  {\frac{{( - 1)^k }}{{k!(k + v)!}}} (bx)^{2k + v} $ is   the Bessel function of the first kind; $L_n^v (\frac{{b^2 }}{a})$ is the Laguerre function. 
And the Laguerre function can be defined in terms of the confluent hypergeometric function $_1 F_1 ( - n,v + 1,\frac{{b^2 }}{a})$ as follow.
$ L_n^v (\frac{{b^2 }}{a}){\rm{ = }}\frac{{(v + n)!}}{{v!n!}}\begin{array}{*{20}c}
   {}  \\
\end{array}_1 F_1 ( - n,v + 1,\frac{{b^2 }}{a})
$
 A: $
\begin{array}{l}
 \int_0^\infty  {\mathop{\rm e}\nolimits}  xp[ - ax^2 ]J_\nu  (2bx)x^{2n + \nu  + 1} dx \\ 
 with\begin{array}{*{20}c}
   {}  \\
\end{array}definition\begin{array}{*{20}c}
   {}  \\
\end{array}of\begin{array}{*{20}c}
   {}  \\
\end{array}J_m (x) \equiv \sum\limits_{{\rm{l = 0}}}^\infty  {\frac{{( - 1)^l }}{{l!(m + l)}}} \left( {\frac{x}{2}} \right)^{2l + m}  \\ 
  = \int_0^\infty  {\mathop{\rm e}\nolimits}  xp[ - ax^2 ]\sum\limits_{{\rm{k = 0}}}^\infty  {\frac{{( - 1)^k }}{{k!(k + \nu )}}} (bx)^{2k + \nu } x^{2n + \nu  + 1} dx \\ 
  = \sum\limits_{{\rm{k = 0}}}^\infty  {\frac{{( - 1)^k }}{{k!(k + \nu )}}} (b)^{2k + \nu } \int_0^\infty  {\mathop{\rm e}\nolimits}  xp( - ax^2 )x^{2k + 2n + 2\nu  + 1} dx \\ 
  = \frac{1}{2}\sum\limits_{{\rm{k = 0}}}^\infty  {\frac{{( - 1)^k }}{{k!(k + \nu )}}} (b)^{2k + \nu } \int_0^\infty  {\mathop{\rm e}\nolimits}  xp( - ax^2 )(x^2 )^{k + n + \nu } d(x^2 ) \\ 
 with\begin{array}{*{20}c}
   {}  \\
\end{array}relation\begin{array}{*{20}c}
   {}  \\
\end{array}of\begin{array}{*{20}c}
   {}  \\
\end{array}\int_0^\infty  {\mathop{\rm e}\nolimits}  xp( - ax)x^m dx = \frac{{\Gamma (m + 1)}}{{a^{m + 1} }} = \frac{{m!}}{{a^{m + 1} }}\begin{array}{*{20}c}
   {}  \\
\end{array}\begin{array}{*{20}c}
   {and\begin{array}{*{20}c}
   {}  \\
\end{array}}  \\
\end{array}\Gamma (m) = (m - 1)! \\ 
  = \frac{1}{2}\sum\limits_{{\rm{k = 0}}}^\infty  {\frac{{( - 1)^k }}{{k!(k + \nu )}}} (b)^{2k + \nu } \frac{{\Gamma (k + n + \nu {\rm{ + }}1)}}{{a^{k + n + \nu {\rm{ + }}1} }} \\ 
 {\rm{ = }}\frac{1}{2}\frac{{b^\nu  }}{{a^{n + \nu {\rm{ + }}1} }}\sum\limits_{{\rm{k = 0}}}^\infty  {\frac{1}{{k!}}\frac{{\Gamma (k + n + \nu {\rm{ + }}1)}}{{\Gamma (k + \nu  + 1)}}( - \frac{{b^2 }}{a}} )^k  \\ 
 {\rm{ = }}\frac{1}{2}\frac{{b^\nu  }}{{a^{n + \nu {\rm{ + }}1} }}\frac{{\Gamma (n + \nu {\rm{ + }}1)}}{{\Gamma (\nu {\rm{ + }}1)}}\sum\limits_{{\rm{k = 0}}}^\infty  {\frac{1}{{k!}}\frac{{\Gamma (n + \nu {\rm{ + }}1 + k)}}{{\Gamma (n + \nu {\rm{ + }}1)}}\frac{{\Gamma (\nu {\rm{ + }}1)}}{{\Gamma (\nu  + 1 + k)}}( - \frac{{b^2 }}{a}} )^k  \\ 
 with\begin{array}{*{20}c}
   {}  \\
\end{array}definition\begin{array}{*{20}c}
   {}  \\
\end{array}of\begin{array}{*{20}c}
   {}  \\
\end{array}_1 F_1 (a,b,z) \equiv 1 + \frac{a}{b}z + \frac{{a(a + 1)}}{{b(b + 1)}}\frac{{z^2 }}{{2!}} = \sum\limits_{{\rm{n = 0}}}^\infty  {\frac{{z^n }}{{n!}}\frac{{(a)^{(n)} }}{{(b)^{(n)} }}}  = \sum\limits_{{\rm{n = 0}}}^\infty  {\frac{{z^n }}{{n!}}\frac{{\Gamma (a + n)}}{{\Gamma (a)}}} \frac{{\Gamma (b)}}{{\Gamma (b + n)}} \\ 
 {\rm{ = }}\frac{1}{2}\frac{{b^\nu  }}{{a^{n + \nu {\rm{ + }}1} }}\frac{{\Gamma (n + \nu {\rm{ + }}1)}}{{\Gamma (\nu {\rm{ + }}1)}}\begin{array}{*{20}c}
   {}  \\
\end{array}_1 F_1 (n + \nu {\rm{ + }}1,\nu  + 1, - \frac{{b^2 }}{a}) \\ 
 with\begin{array}{*{20}c}
   {}  \\
\end{array}relation\begin{array}{*{20}c}
   {}  \\
\end{array}of\begin{array}{*{20}c}
   {}  \\
\end{array}\begin{array}{*{20}c}
   {}  \\
\end{array}_1 F_1 (b - a,b, - x) = \exp ( - x)\begin{array}{*{20}c}
   {}  \\
\end{array}_1 F_1 (a,b,x) \\ 
 {\rm{ = }}\frac{1}{2}b^\nu  a^{ - n - \nu {\rm{ - }}1} \frac{{(n + \nu )!}}{{\nu ! }}\exp ( - \frac{{b^2 }}{a})\begin{array}{*{20}c}
   {}  \\
\end{array}_1 F_1 ( - n,\nu  + 1,\frac{{b^2 }}{a}) \\ 
 with\begin{array}{*{20}c}
   {}  \\
\end{array}relation\begin{array}{*{20}c}
   {}  \\
\end{array}of\begin{array}{*{20}c}
   {}  \\
\end{array}L_n^v (\frac{{b^2 }}{a}){\rm{ = }}\frac{{(v + n)!}}{{v!n!}}\begin{array}{*{20}c}
   {}  \\
\end{array}_1 F_1 ( - n,v + 1,\frac{{b^2 }}{a}) \\ 
 {\rm{ = }}\frac{{n!}}{2}b^\nu  a^{ - n - \nu {\rm{ - }}1} \exp ( - \frac{{b^2 }}{a})L_n^v (\frac{{b^2 }}{a}) \\ 
 \end{array}
$
In summary, the following equation was proved:
$
\int_0^\infty  {\mathop{\rm e}\nolimits}  xp[ - ax^2 ]J_\nu  (2bx)x^{2n + \nu  + 1} dx = \frac{{n!}}{2}b^\nu  a^{ - n - \nu {\rm{ - }}1} \exp ( - \frac{{b^2 }}{a})L_n^v (\frac{{b^2 }}{a})
$
