# Definite integral of polynomial times exponential times hypergeometric function of imaginary argument

How would one deal with such an integral?

$$I(k)\equiv \int_0^\infty r^n e^{-r(1+\mu)} e^{-{\mathrm i} kr}\:{}_1F_1({\mathrm i}/k+1;2;2{\mathrm i} kr) \, \mathrm{d} r$$

Here $n\in\{0,1\}$, $\mu\in \mathbb{R_0^+}$, $k>0$ is a real constant, $\;{}_1F_1$ is the confluent hypergeometric function (${\mathrm i}$ ist the imaginary unit).

Actually, if one tries with Mathematica, one gets a conditional expression with $\operatorname{Im}(k) > 0$ (see bbgodfrey's answer). If we assume $k$ to be real, Mathematica says that the integral does not converge, see this post of mine (derived from this question) in the Mathematica forum.

But if we insert this integral into the following integral:

$$\int_0^\infty\frac{k^2 e^{\pi/k}}{1+k^2}|\Gamma{(1-{\mathrm i}/k})|^2|I(k)|^2\mathrm{d} k,$$

the full double integral converges and Mathematica can handle it in less than one minute if numerical values of $n$ and $\mu$ are chosen. For n=1 and μ=0 it gives 0.0650882 (confirming the result of another perturbative method, see the motivation below). How to make sense of the behavior of the integrals and how to treat them correctly?

Motivation - where does it come from?

The integral arises from a problem in quantum mechanics. In more detail, it is the contribution of continuous states to second order correction to ground state energy (in the framework of Rayleigh-Schroedinger perturbation theory). The unperturbed system is the ground state of the hydrogen atom with radial function proportional to $R_0=e^{-r}$, the perturbation is the screened Coulomb potential (sometimes called the Yukawa potential) $H_p=e^{-\mu r}/r$. The expression for spherically symmetric continuous states of the hydrogen atom is $R_{k,l=0}(r)=\sqrt{\frac2{\pi}}k e^{\pi/2k} |\Gamma{(1-{\mathrm i}/k})| e^{-{\mathrm i} kr} {}_1F_1({\mathrm i}/k+1,2,2{\mathrm i} kr)$. Putting this into the expression for second order correction (from continuous states, apart from this, there is also the contribution from bound states):

$$E_0^{(2)}(continuum)= \int_0^\infty \frac{\left|\int_0^\infty r^2 R_0 H_p R_{k,l=0}\,\mathrm{d} r\right|^2}{E_0-E_k}\,\mathrm{d} k$$

gives the above integral (ground state energy $E_0=-1$, energy of continuous states $E_k=k^2$).

Update: see Jens' answer here: https://mathematica.stackexchange.com/a/78320/22001

• @Alex R. But the first term involves $\int_0^\infty\frac{e^{-n r}}{r}dr$ which is divergent. – Harry Peter Mar 11 '15 at 13:08
• Alex R. claimed to express $_1F_1(i/k+1;2;2i kr)$ as power series form and integrate termwisely. – Harry Peter Mar 17 '15 at 12:26
• Can you include a bit more detail on the QM problem which motivates this question? That could help us to see what approaches are most likely to bear fruit. – Semiclassical Mar 25 '15 at 14:42
• You have $n$ in two places, are they the same? (I wonder because you say that $n\in\{0,1\}$ and $n\in\mathbb{N}$; which is it?) – 2012rcampion Mar 25 '15 at 16:57
• My Mathematica fails to evaluate $I(k)$, given your assumptions for the parameters. It does not evaluate even with $n=\mu=0$. Are you sure this worked? – Jinxed Mar 25 '15 at 18:03

## 1 Answer

With no conditions applied to the parameters, Mathematica gives for the integral,

((-I)*k)^(-1 - n)*(-(k/(k - I*(1 + mu))))^(1 + n)*Gamma[1 + n]*
Hypergeometric2F1[(I + k)/k, 1 + n, 2, (2*k)/(k - I*(1 + mu))]


but as part of a ConditionalExpresson with conditions that are not satisfied for k real. Nonetheless, the expression seems well behaved there and gives the same numerical results as sample numerical integrations.