I am trying to find the asymptotic expansion of $_1F_1\left(-m;\frac{1}{2};-\frac{1}{2}\right)$ for large $m$ where $_1F_1\left(a;b;z\right)$ is the Kummer confluent hypergeometric function, also denoted as $M(a,b,z)$ in Chapter 13 Confluent Hypergeometric Functions of the Digital Library of Mathematical Functions (http://dlmf.nist.gov/13). In the standard notation I am interested in the limit $a \to -\infty$ with the following values fixed $b=1/2$ and $z=-1/2$. In section 13.8 of DLFM they give approximations for:

  1. $a \to \infty$ and $b \leq 1$ (Eq. 13.8.8)

  2. $a \to -\infty$ and $b \geq 1$ (Eq. 13.8.9)

  3. $a \to -\infty$ and $(b-1)/|a|$ positive

  4. Finally $a \pm \infty$ and $\text{ph}(a) \leq \pi -\delta$ i.e. $a$ cannot be purely real negative. (See Eq. 13.8.13)

As you can see none of these cases apply to my problem. Any suggestions on how to get the scaling?


Don't know if the question is still relevant, but the asymptotics can be obtained as follows. Use the identity $${_1F_1}\left(-m;\frac{1}{2};-\frac{1}{2}\right)= e^{-1/2}\,{_1F_1}\left(m+\frac{1}{2};\frac{1}{2};\frac{1}{2}\right)$$ Then use the integral representation $${_1F_1}\left(m+\frac{1}{2};\frac{1}{2};\frac{1}{2}\right)= \frac{1}{\Gamma\left(m+\frac{1}{2}\right)} \int_{0}^{\infty}e^{-t}t^{m-1/2}\,{_0F_1}\left(;\frac{1}{2};\frac{t}{2}\right)dt= \\\frac{1}{\Gamma\left(m+\frac{1}{2}\right)} \int_{0}^{\infty}e^{-t}t^{m-1/2}\cosh(\sqrt{2\,t})dt$$ The exponent is $-t+(m-1/2)\ln(t)+\sqrt{2\,t}$; the other term coming from $\cosh$ contributes subdominant terms and can be neglected. Apply (a variation of) Laplace's method to get $${_1F_1}\left(-m;\frac{1}{2};-\frac{1}{2}\right)\sim e^{\sqrt{2\,m}-1/4} \left(\frac{1}{2}+\frac{7\sqrt{2}}{96\sqrt{m}}-\frac{23}{4608m}+\ldots\right)$$

  • $\begingroup$ Don't worry, questions are forever relevant here. Until your answer answers the question and significantly differs from the already posted answers, it will be okay. Welcome on the Math SE! :-) $\endgroup$ – peterh Jan 16 '18 at 17:30
  • $\begingroup$ Dear @Maxim -- Thank you so much. This precisely what I was looking for! $\endgroup$ – Nicolás Quesada Jan 17 '18 at 18:25

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