# Asymptotic Expression involving generalized hypergeometric function

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).

• I think you got this function from another formula with a software, can I ask what is the source expression? – Nosrati Sep 15 '18 at 13:49
• Please see the renew problem @Nosrati – Hui Zhao Sep 15 '18 at 14:44
• @HuiZhao Hey buddy! Can you please provide me some good source to learn about asymptotes in detail? I am searching the web but there is so little content everywhere. Please help .. – Vicrobot Sep 15 '18 at 14:49
• @Vicrobot please review the link the asymptotic expression for a generalized hypergeometric function. I need a similar form for this problem math.stackexchange.com/questions/2902046/… – Hui Zhao Sep 15 '18 at 18:04

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: $$G(\gamma) \sim \cases { \frac {\Gamma(m - k)} {\Gamma(k + 1) \Gamma(m)} \left( \frac {k m \gamma} {\overline \gamma} \right)^k & k < m \\ \frac 1 {k \,\Gamma^2(k)} \left( \frac {k^2 \gamma} {\overline \gamma} \right)^k \ln \overline \gamma & k = m \\ \frac {\Gamma(k - m)} {\Gamma(k) \Gamma(m + 1)} \left( \frac {k m \gamma} {\overline \gamma} \right)^m & k > m }, \quad \overline \gamma \to \infty.$$

• I validated your asymptotic expression in Matlab. Although the two cases ($m<k$ and $m>k$) are matched in high $\overline \gamma$, the case for $m=k$ does not match the exact result. Please double check the case for $m=k$. Thank you! – Hui Zhao Sep 16 '18 at 20:18
• Consider what the next term is for the case $k = m$. In the same way as in your question that I linked to, there is a $\overline \gamma^{\,-k} \ln \overline \gamma$ term and a $\overline \gamma^{\,-k}$ term. $G(\gamma)/(C \,\overline \gamma^{\,-k} \ln \overline \gamma)$ converges to $1$ only as fast as $1/\ln \overline \gamma$ converges to zero. – Maxim Sep 17 '18 at 7:24