Asymptotic form for a generalized hypergeometric function

Question

${}_3{F_2}\left( {m+m_s,{m_s},{m_s};1 + {m_s},1 + {m_s}; - \frac{{{m_s}\overline \gamma }}{{m{\gamma _0}}}} \right)$, where $m$, $m_s$, $\gamma_0$, and $\overline \gamma$ are positive real numbers. When $\overline \gamma$ goes to infinity, can we derive asymptotic form for that generalized hypergeometric function?

Answer 1

The structure of the asymptotic expansion can be seen from the representation of ${_3F_2}$ in terms of the Meijer G-function: $$F(x) = {_3F_2}(m + m_s, m_s, m_s; 1 + m_s, 1 + m_s; -x) = \\ \frac {m_s^2} {\Gamma(m + m_s)} G_{3, 3}^{1, 3} \left( x \middle| {1 - m_s, 1 - m_s, 1 - m - m_s \atop 0, -m_s, -m_s} \right) = \\ \frac {m_s^2} {2 \pi i \Gamma(m + m_s)} \int_\mathcal L \frac {\Gamma(-y) \Gamma(m + m_s + y) \Gamma^2(m_s + y) x^y} {\Gamma^2(1 + m_s + y)} dy.$$ For $x > 1$, the integral is $2 \pi i$ times the sum of the residues at the points $y = -m_s$ and $y = -m -m_s - k$. The leading term in the expansion will be determined by the residue at the double pole $y = -m_s$, giving $$\operatorname{Res}_{y = -m_s} \frac {\Gamma(-y) \Gamma(m + m_s + y) \Gamma^2(m_s + y) x^y} {\Gamma^2(1 + m_s + y)} = \\ \Gamma(m) \Gamma(m_s) x^{-m_s} (\ln x + \psi(m) -\psi(m_s)), \\ F(x) = \frac {m_s} {\binom {m + m_s - 1} {m_s}} x^{-m_s} \ln x + O(x^{-m_s}), \quad x \to \infty.$$