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  • $\begingroup$ Sorry, did you realize that your g0 coincides with the one I found? BTW, using the strategy I outlined yields g1[y_,l0_,l1_,m0_,m1_]=(l0/((1 + l0)*l1* ((E^(1 + l0 + m0)*(2*Pi)^(l0/2))/ (E^(-(1 + y)^2/2))^l0)^ (1 + l0)^(-1)*ProductLog[ (E^((l0*(l1 + m1 - l1*Log[1/(E^((-1 + y)^2/ 2)*Sqrt[2*Pi])]))/ ((1 + l0)*l1))*l0)/ ((1 + l0)*l1* ((E^(1 + l0 + m0)*(2*Pi)^ (l0/2))/(E^(-(1 + y)^2/ 2))^l0)^(1 + l0)^ (-1))]))^(1 + l0^(-1)) $\endgroup$ Commented Apr 21, 2017 at 15:17
  • $\begingroup$ @marmot Thanks for pointing out that the solutions agree. I had not realized that before. And thanks for giving an expression for $g_1$. After plotting $g_0$ for some values of the parameters, I am thinking there may be ways to bound the solution for the parameters. $\endgroup$ Commented Apr 21, 2017 at 19:01
  • $\begingroup$ Thanks for checking. So, according to Seyhmus Güngören, your solution must be wrong ;-) (I'm kidding, both solutions are correct, I believe.) I also have problems in doing the numerical integration, but after Seyhmus Güngören's reaction my motivation to fix this dropped. $\endgroup$ Commented Apr 21, 2017 at 19:07