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I'd like to solve the following equation numerically for $a_1$, $a_2$, $\alpha_1$ and $\alpha_2$:

$\displaystyle\int_0^{\infty}\Big[L_1\Big(\frac{k^2}{2}\Big)\Big]^2e^{-k^2}L_m(k^2)\ dk=$

$\displaystyle\int_0^{\infty}\bigg(1+\frac{k\cdot a_1}{2\alpha_1}e^{-\frac{k^2}{4\alpha_1}}-\frac{k\cdot a_2}{8\alpha_2}e^{-\frac{k^2}{4\alpha_2}}\Big(\frac{k^2}{\alpha_2}-4\Big)\bigg)e^{-k^2}L_m(k^2)\ dk\ \ \ \text{for}\ \ m=1,3,5,7$

where $L_m(x)$ is the Laguerre polynomial.

Is there a nice way to do this in mathematica? Naively I've tried FindRoot and Solve, but run into all sorts of trouble.

I guess looking into some approximations could do it, just wanted to check if there is an (more or less) obvious way first.

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Please provide the Mathematica code so that it can be easily copied. –  ybeltukov Sep 24 '13 at 20:14
This integral can be easily calculated by Mathematica. Then you can apply Solve as usual. –  ybeltukov Sep 24 '13 at 20:44
Huh! Why didn't I think of that. I don't have enough rep to vote up your comments now, I'll try to remember later. Thanks! –  jorgen Sep 24 '13 at 21:31
@jorgen Are you sure the equations are written correctly? Are there any assumptions for the coefficients? Integrals could be computed easily, but Solve could not find solutions for over 30 minutes. Once you write $α2$, in another place $α_{2}$. –  Artes Sep 24 '13 at 21:33
@Artes Yeah pretty sure, except it said $\alpha2$ instead of $\alpha_2$ one place, fixed now. It is possible I must use an approximation instead of exact equality; I'll look into it. –  jorgen Sep 24 '13 at 21:38

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