Table of Integrals, Series, and Products, 8th Edition

${\displaystyle {\int }_0^{\infty }{x}^{\mu -v+1+2n}{J}_{\mu }(ax)J_{\mu }(bx)\frac{\mathrm{d}x}{x^2+c^2}}={(-1)}^n{c}^{\mu -v+2n}{I}_v(bc){\mathit{K}}_{\mu }(ac)$

  ET II 49(15)

$[b>0,\text{?}\text{?}a>b,\text{?}\text{?}Rec>0,\text{?}Recv-2n+2>Re\mu >-n-1,\text{?}\text{?}n\ge 0\text{?}\text{an}\text{?}\text{integer}]$

6.578

1. 

$\begin{array}{cc}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{x}^{\varrho -1}{J}_{\lambda }(ax)J_{\mu }(bx)J_v(cx)\mathrm{d}x& =\frac{2^{\varrho -1}{a}^{\lambda }{b}^{\mu }{c}^{-\lambda -\mu -\varrho }\Gamma (\frac{\lambda +\mu +v+\varrho }{2})}{\Gamma (\lambda +1)\Gamma (\mu +1)\text{?}\Gamma \text{?}(1-\frac{\lambda +\mu +v+\varrho }{2})}\\ \times F_4(\frac{\lambda +\mu -v+e}{2},\frac{\lambda +\mu +v+\varrho }{2};\lambda +1,\mu +1;\frac{a^2}{c^2},\frac{b^2}{c^2})\end{array}$

  ET II 351(9)

$[Re(\lambda +\mu +v+\varrho )>0,\text{?}Re\varrho <\frac{5}{2},\text{?}a>0,\text{?}b>0,\text{?}c>0,\text{?}c>a+b]$

2. 

$\begin{array}{c}\underset{0}{\overset{\infty }{{\displaystyle \int }}}{x}^{\varrho -1}{J}_{\lambda }(ax)J_{\mu }(bx\end{array}$