Working on a problem in spectral theory, I need to study the asymptotics of a confluent hypergeometric function (here $(a)_0=1$ and $(a)_s=a(a+1)\cdots(a+s-1)$ denote the Pochhammer symbol) $$ \mathbf{M}(a,b,z)=\sum_{s=0}^{+\infty}\frac{(a)_s}{\Gamma(b+s)s!}z^s,\quad\text{as}\ z\to+\infty. $$ In my case $-1<a<0$ and $b=1$, and I'm only interested in real $z$.

I had a look in Abramowitz–Stegun (13.5.1, where $M(a,b,z)=\Gamma(b)\mathbf{M}(a,b,z)$), and found that, as $z\to+\infty$, we have the expansion $$ \mathbf{M}(a,1,z)\sim \frac{e^{i\pi a}}{\Gamma(1-a)}z^{-a}\sum_{s=0}^{+\infty}\frac{\bigl((a)_s\bigr)^2}{s!(-z)^s} + \frac{e^z z^{a-1}}{\Gamma(a)}\sum_{s=0}^{+\infty}\frac{\bigl((1-a)_s\bigr)^2}{s!z^s}. $$ What is worrying me is the factor $e^{i\pi a}$ in the first term. It is complex (in fact, non-real), and everything else in the expansion is real for positive $z$. Also, from the definition of $\mathbf{M}$ we see that it should be real for positive $z$. I have also had a look in 13.7.2, where the same expansion is given. It is also the same in the book Asymptotics and special functions by Frank Olver, and I get the same from Mathematica. Thus, I believe that the expansion above is correct.

In the asymptotic expansion above the term with the $e^{i\pi a}$ factor is small in comparison with the second one. In fact, some sources hint that it can be neglected (compare 13.7.1). As it happens, I want to keep that term, even if it is small. Thus, I think I can state my questions as follows:

  1. Why is the real-valued function having complex terms in its asymptotic expansion?
  2. I'm only considering real $z$. Will the expansion of $\mathbf{M}(a,1,z)$ above still be valid if I replace $e^{i\pi a}$ with its real part, $\cos(\pi a)$?
  3. Could it be that the imaginary part somehow cancels? (I don't see how it could.)

When z is real and tends to $\infty$ (that is, $z>0$) the second part of this double series is dominant, that is, is exponentially bigger that the first part and then, in asymptotic sense, you can avoid it to obtain

$$\mathbf{M}(a,1,z)\sim \frac{e^z z^{a-1}}{\Gamma(a)}\sum_{s=0}^{+\infty}\frac{\bigl((1-a)_s\bigr)^2}{s!z^s}$$

so, the complex terms in its asymptotic expansion dissapear. Note that when $z$ is real and negative, that is, $z\to-\infty$, is the first part that is dominant and taking account that $z^{-a}=e^{i\,\pi\,a}(-z)^{-a}$, the imaginary part also dissapear

$$\mathbf{M}(a,1,z)\sim \frac{1}{\Gamma(1-a)}(-z)^{-a}\sum_{s=0}^{+\infty}\frac{\bigl((a)_s\bigr)^2}{s!(-z)^s}$$

where here $-z$ is positive.

  • $\begingroup$ Thank you for your answer. What you write is clear to me, but I cannot throw away the smaller part. I should perhaps have been more precise in my question. I know that for $-1<a<0$ there is a unique zero of $a\mapsto \mathbf{M}(a,1,z)$ as $z$ is large (positive). I want to find the asymptotics of this zero as $z\to+\infty$. The only term that can help me is $1/\Gamma(a)$, and as $a\to 0^-$ this term tends to zero. However, to get the correct asymptotics ($a\sim -z e^{-z}$) I need the first term from the non-dominate part of the expansion. $\endgroup$ – mickep Apr 4 '16 at 12:09
  • $\begingroup$ Yes, maybe you must be more explicit in your question and explain better this last purpose. $\endgroup$ – ppooppii Apr 4 '16 at 12:48

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