Asymptotic Expression involving generalized hypergeometric function

Question

What is the asymptotic expression for $F\left( \gamma \right) = \frac{1}{{\Gamma \left( m \right)\Gamma \left( k \right)}}\left[ {\frac{1}{k}{{\left( {\frac{{km\gamma }}{{\overline \gamma }}} \right)}^k}\Gamma \left( { - k + m} \right)\frac{{_1{F_2}\left( {k;1 + k,1 + k - m;\frac{{km\gamma }}{{\overline \gamma }}} \right)}}{{\Gamma \left( {1 + k} \right)\Gamma \left( {1 + k - m} \right)}} + \frac{1}{m}{{\left( {\frac{{km\gamma }}{{\overline \gamma }}} \right)}^m}\Gamma \left( {k - m} \right)\frac{{{}_1{F_2}\left( {m;1 + m,1 - k + m;\frac{{km\gamma }}{{\overline \gamma }}} \right)}}{{\Gamma \left( {1 + m} \right)\Gamma \left( {1 - k + m} \right)}}} \right]$, when $\overline \gamma \to +\infty$, where $m$, $k$, $\gamma$ and $\overline \gamma$ are positive real numbers, and $\Gamma(\cdot)$ denotes the Gamma function, and ${}_1 F_2\cdot (\cdot)$ denotes the Generalized hypergeometric function (https://en.wikipedia.org/wiki/Generalized_hypergeometric_function). There are three cases for this asymptotic expression, i.e., $m<k$, $m>k$ and $m=k$. Actually, $F(\gamma)$ is the cumulative density function of random variable $\gamma$, which is derived from its probability density function, given by $f\left( \gamma \right) = \frac{2}{{\Gamma \left( m \right)\Gamma \left( k \right)}}{\left( {\frac{{km}}{{\overline \gamma }}} \right)^{\frac{{\left( {k + m} \right)}}{2}}}{\gamma ^{\frac{{\left( {k + m - 2} \right)}}{2}}}{K_{k - m}}\left( {2\sqrt {\frac{{km\gamma }}{{\overline \gamma }}} } \right)$, where $K_{k-m}(\cdot)$ denotes the Modified Bessel K functions (https://en.wikipedia.org/wiki/Bessel_function).

Answer 1

Your $F(\gamma)$ is not an antiderivative of $f(\gamma)$ because the factors in front of ${_1F_2}$ are off. In any case, a representation of $F(\gamma)$ in terms of ${_1F_2}$ for integral $k, m$ will be valid only as a limit. The reason is that $$f(\gamma) = \frac 1 {\Gamma(k) \Gamma(m) \gamma} G_{0, 2}^{2, 0} \left( \frac {k m \gamma} {\overline \gamma} \middle| {- \atop k, m } \right), \\ G(\gamma) = \int_0^\gamma f(\xi) d\xi = \frac 1 {\Gamma(k) \Gamma(m)} G_{1, 3}^{2, 1} \left( \frac {k m \gamma} {\overline \gamma} \middle| {1 \atop k, m, 0} \right),$$ and the sum of the residues has to be taken over an infinite sequence of double poles, giving an infinite sum of polygamma functions instead of a sum of gamma functions. The leading term can be found in the same way as here: $$G(\gamma) \sim \cases { \frac {\Gamma(m - k)} {\Gamma(k + 1) \Gamma(m)} \left( \frac {k m \gamma} {\overline \gamma} \right)^k & $k < m$ \\ \frac 1 {k \,\Gamma^2(k)} \left( \frac {k^2 \gamma} {\overline \gamma} \right)^k \ln \overline \gamma & $k = m$ \\ \frac {\Gamma(k - m)} {\Gamma(k) \Gamma(m + 1)} \left( \frac {k m \gamma} {\overline \gamma} \right)^m & $k > m$ }, \quad \overline \gamma \to \infty.$$