Solve the integral
$$ \begin{aligned} I &= \int_{X_{0\ell}} \int_{Y_{0\ell}} \left[\sum_{\ell=1}^{m_{1}}\left(x_{0 \ell}^{2}+y_{0 \ell}^{2}\right)\right]^{k_{1}} \left[\sum_{\ell=1}^{m_{2}}\left(x_{0 \ell}^{2}+y_{0 \ell}^{2}\right)\right]^{k_{2}} \\ & \times exp\left[-h_{1}\sum_{\ell=1}^{m_{1}}\left(x_{0 \ell}^{2}+y_{0 \ell}^{2}\right)\right]exp\left[-h_{2}\sum_{\ell=1}^{m_{1}}\left(x_{0 \ell}^{2}+y_{0 \ell}^{2}\right)\right] d{X_{0\ell}} d{Y_{0\ell}} \end{aligned} $$ The integration is over $-\infty<{X_{0\ell}},{Y_{0\ell}}<\infty$, where ($k_{1}$ , $k_{2}$) are real positive integers and ($h_{1}$,$h_{2}$) are real positive numbers. A special case of the integral is when $m_{1}=m_{2}=m$, then the integral is reduced to $$ \begin{aligned} I &= \int_{X_{0\ell}} \int_{Y_{0\ell}} \left[\sum_{\ell=1}^{m}\left(x_{0 \ell}^{2}+y_{0 \ell}^{2}\right)\right]^{k_{1}+k_{2}} exp\left[-(h_{1}+h_{2})\sum_{\ell=1}^{m}\left(x_{0 \ell}^{2}+y_{0 \ell}^{2}\right)\right] d{X_{0\ell}} d{Y_{0\ell}} \end{aligned} $$ The above integral can be evaluated easily by doing a change the variable. We use the hyper-spherical coordinates $$ \begin{aligned} x_{01} &= t\operatorname{cos}\left(\theta_{1}\right) \\ y_{01} &=t\operatorname{sin}\left(\theta_{1}\right) \cos \left(\theta_{2}\right) \\ x_{02} &=t\operatorname{sin}\left(\theta_{1}\right) \sin \left(\theta_{2}\right) \cos \left(\theta_{3}\right) \\ & \vdots \\ x_{0 m_{k}} &=t\operatorname{sin}\left(\theta_{1}\right) \sin \left(\theta_{2}\right) \cdots \cos \left(\theta_{2 m_{k}-1}\right) \\ y_{0 m_{k}} &=t\operatorname{sin}\left(\theta_{1}\right) \sin \left(\theta_{2}\right) \cdots \sin \left(\theta_{2 m_{k}-1}\right) \end{aligned} $$ The Jacobian will be $$ |J|=t^{2 m-1} \sin ^{2 m-2}\left(\theta_{1}\right) \sin ^{2 m-3}\left(\theta_{2}\right) \cdots \sin \left(\theta_{2 m-2}\right) $$ Then the integral is changed to $$ \begin{aligned} I=\int_{t=0}^{\infty} \int_{\theta_{1}=0}^{\pi} \int_{\theta_{2}=0}^{\pi} \cdots \int_{\theta_{2m-1}=0}^{2\pi}& t^{2 m-1} \sin ^{2 m-2}\left(\theta_{1}\right) \sin ^{2 m-3}\left(\theta_{2}\right) \cdots \sin \left(\theta_{2 m-2}\right)\\ & \times t^{2(k_{1}+k_{2})} \:\:e^{-(h_{1}+h_{2})t^2} \:\: d \theta_{1} d \theta_{2} \cdots d \theta_{2 m-1} \:\: d t \end{aligned} $$ The integral over angles can be evaluated by mathematical induction, and by letting $t^2=u$, then $$ \begin{aligned} I=\frac{\pi^m}{\Gamma(m)}\int_{u=0}^{\infty} u^{(k_{1}+k_{2}+m-1)} \:\:e^{-(h_{1}+h_{2})u} \:\: d u \end{aligned} $$ $$ \begin{aligned} =\frac{\pi^m}{\Gamma(m)} \Gamma(k_{1}+k_{2}+m) \:\:(h_{1}+h_{2})^{-(k_{1}+k_{2}+m)} \end{aligned} $$ Now, can we solve it for $m_{1} \neq m_{2}$, supposing that $m_{2}>m_{1}$?