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I have the following nasty expression that I would like to expand in powers of $\frac{1}{N}$:

\begin{align} \frac{2^{\frac{3}{2}} 3^{\frac{1}{2}} \Biggl[ \sqrt{u} \cdot \Gamma\left(\frac{2+N}{4}\right) \cdot {}_1F_1 \left( \frac{2+N}{4},\frac{1}{2},\frac{3r^2}{2u} \right) -\sqrt{6} r \cdot \Gamma \left( \frac{4+N}{4} \right) \cdot {}_1F_1 \left( \frac{4+N}{4},\frac{3}{2},\frac{3r^2}{2u} \right) \Biggr] }{N \cdot u^{\frac{1}{2}} \Biggl[ \sqrt{u} \cdot \Gamma\left(\frac{N}{4}\right) \cdot {}_1F_1 \left( \frac{N}{4},\frac{1}{2},\frac{3r^2}{2u} \right) -\sqrt{6} r \cdot \Gamma \left( \frac{2+N}{4} \right) \cdot {}_1F_1 \left( \frac{2+N}{4},\frac{3}{2},\frac{3r^2}{2u} \right) \Biggr]} \end{align}

where you can assume that $N \in \mathbb{N}$ (but could be analytically continued to $\mathbb{R}^+$), $u \in \mathbb{R}^+$, and $r \in \mathbb{R}^+$. Furthermore, ${}_1F_1$ is the confluent hypergeometric function sometimes written as $M(a,b,z)$.

Using a different route I have obtained a value for the limit $N \to \infty$, but I'd like to a) reproduce this result using the above expression and b) find the $O\left(\frac{1}{N}\right)$ corrections. So far I have tried numerous identities from the NIST Handbook of Mathematical Functions, but I simply seem to lack the experience to make real progress. If anyone knows a solution or has an idea of how to proceed next, I'd greatly appreciate their help.

With best regards,


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1 Answer 1

Perhaps not a big help: For ${}_1F_1((2+N)/4,3/2,3r^2/2u)$ we want asymptotics of ${}_1F_1(1/2+x,3/2,a)$ as $x \to 0$. Using the series, I get $$ {}_1F_1\left(\frac{1}{2}+x,\frac{3}{2},a\right)= \frac{\sqrt{\pi} \mathrm{erf} (i \sqrt{a})}{2 \sqrt{-a}} + \sum_{k = 0}^{\infty} \frac{\Bigl(\Psi \Bigl(\frac{1}{2} + k\Bigr) + \gamma + 2 \operatorname{ln} (2)\Bigr) a^{k}}{(1 + 2 k) k!} x + \operatorname{O} \bigl(x^{2}\bigr) $$

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