Imaginary part in asymptotics of hypergeometric function

Question

I am working with the function

$$ _2F_3\left(n+\frac{1}{2},n+\frac{3}{2};n,n+\frac{5}{2},2 n+1;x^2\right).$$

Evaluating the numerical value of this or plotting it indicates that this is always real for integer $n$. However, when I use Mathematica to find the expansion of this function near $x=\infty$ using Series[HypergeometricPFQ[{1/2 + n, 3/2 + n}, {n, 5/2 + n, 1 + 2 n}, x^2], {x, \[Infinity], 2}], the expansion has imaginary parts.

Why does this happen? How can I find the asymptotic behavior of this function? Thanks a lot!

Answer 1

I am not sure that, for general $n$, you could obtain this expansion even if you add Assumptions -> n > 0 && n Integer

For $n=3$ for example, your command gives $$\left(\frac{1728 i}{\pi x^7}+O\left(\left(\frac{1}{x}\right)^8\right)\right) \exp \left(-2 x-\frac{7 i \pi }{2}+O\left(\left(\frac{1}{x}\right)^3\right)\right)+\left( \frac{1728 i}{\pi x^7}+O\left(\left(\frac{1}{x}\right)^8\right)\right) \exp \left(2 x+\frac{7 i \pi }{2}+O\left(\left(\frac{1}{x}\right)^3\right)\right)+O\left( \left(\frac{1}{x}\right)^7\right)$$ Now, use FullSimplify[Normal[%]] which will produce $$\frac{6912 \sinh (x) \cosh (x)}{\pi x^7}$$ which can simplify.

In fact, $$\, _2F_3\left(n+\frac{1}{2},n+\frac{3}{2};n,n+\frac{5}{2},2 n+1;x^2\right)=a_n \, \frac{\sinh(2x) }{\pi \, x^{2 n+1} }+\cdots$$ where the $a_n$ form the (unknown ?) sequence $$\{10,112,3456,202752,19169280,2654208000,505361203200,\cdots\}$$

Edit (to avoid loosing @Gary's comment)

$$a_n=2^{2n - 1}\, (2n + 3)\,n!\,(n - 1)!$$