La suma se puede expresar también en términos de la Regularización de la Gamma Incompleta de la función ($Q(a,z)$).
De hecho, la premisa de que para que una función general de un entero $f(k)$ hemos
$$
\begin{gathered}
\frac{{\left( {i^{\,k} + \left( { - i} \right)^{\,k} } \right)}}
{2}f(k) = \left( {\frac{{e^{\,i\,k\frac{\pi }
{2}} + e^{\, - \,i\,k\frac{\pi }
{2}} }}
{2}} \right)f(k) = \cos \left( {k\frac{\pi }
{2}} \right)f(k) = \hfill \\
= \left[ {k = 2j} \right]\left( { - 1} \right)^{\,j} f(2j)\quad \left| {\;k,j\; \in \;\;\mathbb{Z}\,} \right. \hfill \\
\end{reunieron}
$$
y que el Menor Gamma Incompleta de la función puede ser expresada como:
$$
\begin{gathered}
\gamma (s,z) = \int_{\,0\,}^{\,z\,} {t^{\,s - 1} \;e^{\, - \,t} dt} = \hfill \\
= z^{\,s} \;e^{\,\, - z} \;\Gamma (s)\;\sum\limits_{0\, \leqslant \,k} {\frac{{z^{\,k} }}
{{\Gamma (s + k + 1)}}} = \frac{{z^{\,s} \;e^{\,\, - z} }}
{s}\sum\limits_{0\, \leqslant \,k} {\frac{{z^{\,k} }}
{{\left( {s + 1} \right)^{\,\overline {\,k\,} } \,}}} = \hfill \\
= z^{\,s} \;e^{\,\, - z} \sum\limits_{0\, \leqslant \,k} {\frac{{z^{\,k} }}
{{s^{\,\overline {\,k + 1\,} } \,}}} = z^{\,s} \sum\limits_{0\, \leqslant \,j} {\frac{{\left( { - 1} \right)^{\,j} }}
{{\,\left( {s + j} \right)}}\frac{{z^{\,j} }}
{{j!}}} \hfill \\
\end{reunieron}
$$
a continuación, podemos escribir
$$
\begin{gathered}
F(x,m) = \sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \frac{{x^{\,2k} }}
{{\left( {2k + m} \right)!}}} = \hfill \\
= \frac{1}
{2}\left( {\sum\limits_{0\, \leqslant \,k} {\frac{{\left( {i\,x} \right)^{\,k} }}
{{\left( {k + m} \right)!}}} + \sum\limits_{0\, \leqslant \,k} {\frac{{\left( { - \,i\,x} \right)^{\,k} }}
{{\left( {k + m} \right)!}}} } \right) = \hfill \\
= \frac{1}
{{2\,\Gamma (m)}}\left( {\frac{{e^{\,\,i\,x} \,\gamma (m,i\,x)}}
{{\left( {i\,x} \right)^{\,m} }} + \frac{{e^{\, - \,i\,x} \,\gamma (m, - i\,x)}}
{{\left( { - i\,x} \right)^{\,m} }}} \right) = \hfill \\
= \frac{1}
{{\left| x \right|^{\,\,m} }}\,\operatorname{Re} \left( {e^{\,\,i\,x} e^{\,\, - \,i\,m\,\left( {sign(x)\pi /2} \right)} \,\frac{{\gamma (m,i\,x)}}
{{\Gamma (m)}}} \right) = \hfill \\
= \frac{1}
{{\left| x \right|^{\,\,m} }}\,\operatorname{Re} \left( {e^{\,\,i\,x} e^{\,\, - \,i\,m\,\left( {sign(x)\pi /2} \right)} \,\left( {1 - Q(m,i\,x)} \right)} \right) \hfill \\
\end{reunieron}
$$
Así, por $x=\pi /2$ tenemos
$$
\begin{gathered}
F(\pi /2,m) = \sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \frac{{\left( {\pi /2} \right)^{\,2k} }}
{{\left( {2k + m} \right)!}}} = \hfill \\
= \left( {\frac{2}
{\pi }} \right)^{\,\,m} \,\operatorname{Re} \left( {e^{\,\, - \,i\,\left( {m - 1} \right)\,\left( {\pi /2} \right)} \,\frac{{\gamma (m,i\,\pi /2)}}
{{\Gamma (m)}}} \right) = \hfill \\
= \left( {\frac{2}
{\pi }} \right)^{\,\,m} \,\operatorname{Re} \left( {e^{\,\, - \,i\,\left( {m - 1} \right)\,\left( {\pi /2} \right)} \,\left( {1 - Q(m,i\,\pi /2)} \right)} \right) \hfill \\
\end{reunieron}
$$
ejemplo con $m=5$
$$
\begin{gathered}
F(\pi /2,5) = \sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \frac{{\left( {\pi /2} \right)^{\,2k} }}
{{\left( {2k + 5} \right)!}}} = \hfill \\
= \left( {\frac{2}
{\pi }} \right)^{\,\,5} \,\operatorname{Re} \left( {e^{\,\, - \,i\,4\,\left( {\pi /2} \right)} \,\left( {1 - Q(5,i\,\pi /2)} \right)} \right) = \hfill \\
= \left( {\frac{2}
{\pi }} \right)^{\,\,5} \,\operatorname{Re} \,\left( {1 - \frac{1}
{{24}}\left( {12\,\pi - \frac{1}
{2}\pi ^{\,3} + i\left( {3\pi ^{\,2} - \frac{1}
{{16}}\pi ^{\,4} - 24} \right)} \right)} \right) = \hfill \\
= \left( {\frac{2}
{\pi }} \right)^{\,\,5} \left( {1 - \,\frac{\pi }
{2} + \frac{1}
{{48}}\pi ^{\,3} } \right) = \left( {\frac{2}
{\pi }} \right)^{\,\,5} - \,\left( {\frac{2}
{\pi }} \right)^{\,\,4} + \frac{1}
{6}\left( {\frac{2}
{\pi }} \right)^{\,2} \hfill \\
\end{reunieron}
$$
que coincide con el valor dado por Zaid Alyafeai,
así como con la fomula indicado por Igor Rivin
$$
\begin{gathered}
F(\pi /2,5) = \sum\limits_{0\, \leqslant \,k} {\left( { - 1} \right)^{\,k} \frac{{\left( {\pi /2} \right)^{\,2k} }}
{{\left( {2k + 5} \right)!}}} = \left( {\frac{2}
{\pi }} \right)^{\,\,5} - \,\left( {\frac{2}
{\pi }} \right)^{\,\,4} + \frac{1}
{6}\left( {\frac{2}
{\pi }} \right)^{\,2} = \hfill \\
= 0.007860176 \cdots = \hfill \\
= \frac{1}
{{5!}}{}_1F_2 \left( {1\;;\;\frac{5}
{2} + \frac{1}
{2},\;\frac{5}
{2} + 1\;;\; - \frac{1}
{4}\left( {\frac{\pi }
{2}} \right)^{\,2} } \right) \hfill \\
\end{reunieron}
$$
y respecto de este último, un equipo de cálculo a través de varios valores de $m$ $x$ se muestra en un partido completo.
