What is the closed form of a certain sum

Question

The following sum appears in a problem of Mathematical Epidemiology:

$$P(m)=\sum _{p=0}^{\infty } \left( \sum _{q=0}^{\infty }2\,{\frac {\min \{ p,q \} {{\rm e}^{-m}} \left( \frac{m}{2} \right) ^{p+q}}{p!\,q !}} \right)$$

Using Maple it is possible to obtain the following curve for $m$ between 0 and 10:

Please let me know what is and how to obtain the closed form. Many thanks

Answer 1

\begin{align*} \sum _{p=0}^{\infty } & \bigg( \sum _{q=0}^{\infty }2\,{\frac {\min\{p,q\} e^{-m}(m/2)^{p+q}}{p!\,q !}} \bigg) \\ &= 2 \sum _{p=0}^\infty \sum _{q=p}^\infty{\frac {p e^{-m}(m/2)^{p+q}}{p!q!}} + 2 \sum _{p=0}^\infty \sum _{q=0}^{p-1} {\frac {q e^{-m}(m/2)^{p+q}}{p!q!}} \\ &= 2 \sum _{p=1}^\infty \sum _{q=p}^\infty{\frac {e^{-m}(m/2)^{p+q}}{(p-1)!q!}} + 2 \sum _{p=0}^\infty \sum _{q=1}^{p-1} {\frac {e^{-m}(m/2)^{p+q}}{p!(q-1)!}} \\ &= 2 \sum _{p=0}^\infty \sum _{q=p+1}^\infty{\frac {e^{-m}(m/2)^{p+1+q}}{p!q!}} + 2 \sum _{p=0}^\infty \sum _{q=0}^{p-2} {\frac {e^{-m}(m/2)^{p+q+1}}{p!q!}} \\ &= 2\sum_{p=0}^\infty \sum_{q=0}^\infty {\frac {e^{-m}(m/2)^{p+q+1}}{p!q!}} -2\frac {e^{-m}(m/2)^{0+0+1}}{0!0!} - 2 \sum _{p=1}^\infty \sum _{q=p-1}^p \frac {e^{-m}(m/2)^{p+q+1}}{p!q!} \\ &= me^{-m} \bigg( \sum_{p=0}^\infty \frac {(m/2)^p}{p!} \bigg) \bigg( \sum_{q=0}^\infty \frac {(m/2)^q}{q!} \bigg) - me^{-m} \\ &\qquad{}- 2 \sum _{p=1}^\infty \bigg( {\frac {e^{-m}(m/2)^{p+(p-1)+1}}{p!(p-1)!}} + {\frac {e^{-m}(m/2)^{p+p+1}}{p!p!}} \bigg) \\ &= me^{-m} e^{m/2}e^{m/2} - me^{-m} - \big( me^{-m} I_1(m) + me^{-m} \big( I_0(m)-1 \big) \big) \\ &= m - me^{-m} \big( I_1(m) + I_0(m) \big), \end{align*} where $I_0$ and $I_1$ are the modified Bessel functions of the first kind.

Answer 2

Using the Procedure of Greg Martin (until the fourth line) and Maple; I am obtaining the alternative closed form

$$P \left( m \right) =m-{{\rm e}^{-m}}m-{\frac {{\rm e}^{-m }{m}^{2} \left( 2+m \right) {_2F_3(1,2+\frac{m}{2};\,2,2,1+\frac{m}{2};\,\frac{m^2}{4})}}{{4}}} $$