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

$\begin{array}{ll}{\displaystyle {\int }_0^{\infty }{x}^{2\mu -\nu }{e}^{-\frac{1}{4}{x}^2}{M}_{k,\mu }\left(\frac{1}{2}{x}^2\right)Y_{\nu }(xy)\text{d}x}\hfill & ={\pi }^{-1}{2}^{\mu +\beta }{y}^{k-\mu -\frac{3}{2}}{e}^{-\frac{1}{4}{y}^2}\mathrm{\Gamma }(2\mu +1)\hfill \\ \times \mathrm{\Gamma }\left(\frac{1}{2}-k-\mu \right)\left\{\cos \left[\left(\nu -2\mu \right)\pi \right]\frac{\mathrm{\Gamma }(2\mu -\nu -1)}{\mathrm{\Gamma }(2\beta +1)}{M}_{\alpha ,\beta }\left(\frac{1}{2}{y}^2\right)-\sin \left[\left(\nu +k-\mu \right)\pi \right]W_{\alpha ,\beta }\pi \left(\frac{1}{2}{y}^2\right)\right\}\hfill \\ \text{?}2\alpha =3\mu -\nu +k+\frac{1}{2},\text{?}\text{?}2\beta =\mu -\nu -k+\frac{1}{2}\hfill \end{array}$

  ET II 116(44)

$\left[y>0,-1<2\mathrm{Re}\mu <\mathrm{Re}(2k+\nu )+\frac{1}{2},\mathrm{Re}(2\mu -\nu )>-1\right]$

14. 

$\begin{array}{ll}{\displaystyle {\int }_0^{\infty }{x}^{2\mu +\nu }{e}^{-\frac{1}{4}{x}^2}{M}_{k,\mu }\left(\frac{1}{2}{x}^2\right)Y_{\nu }(xy)\text{d}x}\hfill & ={\pi }^{-1}{2}^{\mu +\beta }{y}^{k-\mu -\frac{3}{2}}\mathrm{\Gamma }(2\mu +1)\hfill \\ \times \mathrm{\Gamma }\left(\frac{1}{2}-\mu -k\right)e^{-\frac{1}{4}{y}^2}\left\{\cos (2\mu \pi )\frac{\mathrm{\Gamma }(2\mu +\nu +1)}{\mathrm{\Gamma }\left(\mu +\nu -k+\frac{3}{2}\right)}{M}_{\alpha ,\beta }\left(\frac{1}{2}{y}^2\right)+\sin \left[\left(\mu -k\right)\pi \right]W_{\alpha ,\beta }\left(\frac{1}{2}{y}^2\right)\right\}\hfill \\ 2\alpha =3\mu +\nu +k+\frac{1}{2},2\beta =\mu +\nu -k+\frac{1}{2}\hfill \end{array}$

  ET II 116(43)

$\left[y>0,-1<2\mathrm{Re}\mu <\mathrm{Re}(2k-\nu )+\frac{1}{2},\mathrm{Re}(2\mu +\nu )>-1\right]$

15. 

$\begin{array}{l}{\displaystyle {\int }_0^{\infty }{x}^{2\mu +}}\hfill \end{array}$