I am trying to find the definite integral of $$I=\int_{0}^{\infty}dx\left(\frac{\sqrt{n^{2l-1}}M_{n,l+\frac{1}{2}}(\frac{2x}{n})}{\Gamma(2l+2)}\right)x\left(\frac{(2r)^{l_f+1} {}_0F_1(2l_f+2;-2r)}{\sqrt{2}\Gamma(2l_f+2)}\right)$$.

Where $M_{n,l+\frac{1}{2}}(\frac{2x}{n})$ is the WhittakerM function, ${}_0F_1(2l_f+1;-2r)$ is the Hypergeometric0F1 function, $r=a_0*x$, where $a_0$ is the bohr radius in Rydberg units ($a_0=\frac{h^2}{4\pi^2m_ee_0^2}$). I am trying to compute this integral for when $n=2,l=0$ and $l_f=1$ and my input is this:

 x (Sqrt[2^(2*0 - 1)]*WhittakerM[2, 0 + 1/2, (2*x)/2])/
  Gamma[2*0 + 2] ((2*a*x)^(1 + 1) Hypergeometric0F1[2*1 + 2, -2*a*x])/(
  Sqrt[2] Gamma[2*1 + 2]), {x, 0, \[Infinity]}]

But I get

Integrate:Integrate was unable to determine the units of quantities that appear
in the input"

And if I omit the bohr radius the output I get is the input, any ideas on what's going wrong?

  • $\begingroup$ Could you please update your title? I think it'd be better to write something like "Integral involving WhittakerM and Hypergeometric0F1". I mean more beneficial for other users as well :-) $\endgroup$
    – bmf
    Mar 26, 2022 at 18:37
  • 1
    $\begingroup$ @bmf I hope it's fine that I used your suggestion for the new title? $\endgroup$ Mar 26, 2022 at 18:39
  • $\begingroup$ Of course it is. Perfectly fine. Thanks a lot! $\endgroup$
    – bmf
    Mar 26, 2022 at 18:40
  • $\begingroup$ Question: did you try to input only the numerical values for the constants without their units? The units can be restored at the end by means of dimensional analysis $\endgroup$
    – bmf
    Mar 26, 2022 at 18:42
  • $\begingroup$ Sort of? Might have been lazy but I tried just replacing $a$ with an undefined $c$ and adding $Assuming[c>0,....]$ which gave me the same result as if I omitted $a$. $\endgroup$ Mar 26, 2022 at 18:45

1 Answer 1



Integrate[x (Sqrt[2^(2*0 - 1)]*WhittakerM[2, 0 + 1/2, (2*x)/2])/
Gamma[2*0 + 2] ((2*a*x)^(1 + 1) Hypergeometric0F1[2*1 + 2, -2*a*x])/(Sqrt[
   2] Gamma[2*1 + 2]) // FunctionExpand, {x, 0, \[Infinity]}]
(* -256 a^2 (4 + a (-9 + 4 a)) E^(-4 a) *)

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