Consider the Kummer differential equation $$ \frac{d}{dz}\left[z^be^{-z}\frac{dw}{dz}\right]=az^{b-1}e^{-z}w,\quad z\in\mathbb{R}. $$ It is an eigenvalue problem of Sturm-Liouville type with weight function $W(z)=z^{b-1}e^{-z}$. The two linearly independent solutions are the Kummer function $M(a,b,z)$ and the Tricomi function $U(a,b,z)$.

My question is what type of boundary conditions (BC) does one need to impose at $z=\pm\infty$ for the usual completeness relation to hold $$ \int da\,w(a,b,z)w(a,b,z')W(z) =\delta(z-z'). $$ Here $w$ is the correct linear combination of $M$ and $U$ to match the BCs.

For example if we want the eigenfunctions $w$ to decay for $z\to\pm\infty$ then $w=U$ is the only choice, because $U\sim z^{-a}$ for large $z$ ($M$ explodes), and then $a$ has to be positive for this to be a proper decay. So I would naively integrate on $a\in\mathbb{R}_+$ in this case.

I also tried to look for this type of integral in tables but haven't found anything useful.

Does anybody have any ideas how to make this integral precise? Thanks.


What you are looking for is called Weyl's limit point(LP)/limit circle(LC) classification. An end point is LC if both solutions are square integrable (w.r.t . the weight function) and LP otherwise. If both endpoints are LC then you need boundary conditions and you will have a complete set of eigenfunctions. Otherwise the spectrum might have a continuous component and the eigenfunction expansion will be an integral transform. The Kummer equation is LC near $0$ for $0<b<2$ and LP otherwise. $\infty$ is always LP. However, I don't know the precise spectrum of this equation. Near $0$ you can take the Kummer function which is entire with respect to the spectral parameter, hence there is a corresponding integral transform $$ \hat{f}(a) = \int_0^\infty M(a,b,x) f(x) W(x) dx $$ and you can go back via $$ f(x) = \int_{-\infty}^\infty M(a,b,x) \hat{f}(a) d\rho(a) $$ where $\rho$ is the associated singular spectral measure.

For general background concerning LC/LP you could e.g. look at


Depending on your background this might be however quite technical. Moreover, it only covers the case where the left endpoint $0$ is regular. For the general case you will need



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