calculus - A closed form of $sum_{k=1}^inftyfrac{(-1)^{k+1}}{k!}Gamma^2left(frac{k}{2}right)$

I am looking for a closed form of the following series

\begin{equation}
\mathcal{I}=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k!}\Gamma^2\left(\frac{k}{2}\right)

\end{equation}

I have no idea how to answer this question. Wolfram Alpha gives me result:

$$\mathcal{I}\approx2.7415567780803776$$

Could anyone here please help me to obtain the closed form of the series preferably (if possible) with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.

Answer

You can use the identity given by the Euler Beta function
$$\int_{0}^{1}x^{a-1} (1-x)^{b-1} \,dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
to state:
$$S=\sum_{k=1}^{+\infty}\frac{(-1)^{k+1}}{k!}\Gamma(k/2)^2=\sum_{k=1}^{+\infty}\frac{(-1)^{k-1}}{k}\int_{0}^{1}\left(x(1-x)\right)^{k/2-1}\,dx $$
and by switching the series and the integral:
$$ S = \int_{0}^{1}\frac{\log(1+\sqrt{x(1-x)})}{x(1-x)}dx = 2\int_{0}^{1/2}\frac{\log(1+\sqrt{x(1-x)})}{x(1-x)}dx,$$
$$ S = 2\int_{0}^{1/2}\frac{\log(1+\sqrt{1/4-x^2})}{1/4-x^2}dx = 4\int_{0}^{1}\frac{\log(1+\frac{1}{2}\sqrt{1-x^2})}{1-x^2}dx,$$
$$ S = 4\int_{0}^{\pi/2}\frac{\log(1+\frac{1}{2}\sin\theta)}{\sin\theta}d\theta.$$
Now Mathematica gives me $\frac{5\pi^2}{18}$ as an explicit value for the last integral, but probably we are on the wrong path, and we only need to exploit the identity

$$\sum_{k=1}^{+\infty}\frac{1}{k^2\binom{2k}{k}}=\frac{\pi^2}{18}$$
that follows from the Euler acceleration technique applied to the $\zeta(2)$-series. The other "piece" (the $U$-piece in the Marty Cohen's answer) is simply given by the Taylor series of $\arcsin(z)^2$. More details to come.

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As a matter of fact, both approaches lead to an answer.
The (Taylor) series approach, as Bhenni Benghorbal shows below, leads to the identity:
$$\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k!}\Gamma^2\left(\frac{k}{2}\right)x^k= 2 \arcsin \left( x/2 \right) \left(\pi - \arcsin \left( x/2\right) \right),\tag{1}$$
while the integral approach, as Achille Hui pointed out in the comments, leads to:
$$\begin{eqnarray*}\int_{0}^{\pi/2}\frac{\log(1+\frac{1}{2}\sin\theta)}{\sin\theta}\,d\theta&=&\int_{0}^{1}\log\left(1+\frac{t}{1+t^2}\right)\frac{dt}{t}\\&=&\int_{0}^{1}\frac{\log(1-t^3)-\log(1-t)-\log(1+t^2)}{t}\,dt\\&=&\int_{0}^{1}\frac{-\frac{2}{3}\log(1-t)-\frac{1}{2}\log(1+t)}{t}\,dt\\&=&\frac{1}{6}\sum_{k=1}^{+\infty}\frac{4+3(-1)^k}{k^2}=\frac{1}{6}\left(4-\frac{3}{2}\right)\zeta(2)=\frac{5\pi^2}{72}.\end{eqnarray*}\tag{2}$$

Thanks to both since now this answer may become a reference both for integral-log-ish problems like $(2)$ and for $\Gamma^2$-series like $(1)$.

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Update 14-06-2016. I just discovered that this problem can also be solved by computing
$$ \int_{-1}^{1} x^n\, P_n(x)\,dx, $$
where $P_n$ is a Legendre polynomial, through Bonnet's recursion formula or Rodrigues' formula. Really interesting.