Is the generalized hypergeometric function ${}_1F_2\bigl(1;a,a+\frac12;-x^2\bigr)$ for $a>-1$ and $x>0$ an elementary function?
How about the positivity, monotonicity, and convexity of the generalized hypergeometric function ${}_1F_2\bigl(1; a, a+\frac{1}{2}; -x^2\bigr)$ in $x>0$ for $a\ge-1$?
Take $a=1/4$.
Maple says
$$
{}_1F_2\left(1;\frac14,\frac34;-x^2\right) =
1-2\,\sqrt {x}\sqrt {\pi}\sin \left( 2\,x \right)\, {C}\!
\left( 2\,{\frac {\sqrt {x}}{\sqrt {\pi}}} \right) +2\,\sqrt {x}
\sqrt {\pi}\cos \left( 2\,x \right)\, {S}\! \left( 2\,{\frac {
\sqrt {x}}{\sqrt {\pi}}} \right)
$$
in terms of the Fresnel integrals $S$ and $C$.
Now, $S$ and $C$ are not elementary. So I guess this combination is also not elementary.
Simplified ... $$ {}_1F_2\left(1;\frac14,\frac34;-x^2\right) = 1+2\sqrt {x}\int_{-x}^{0}\!{\frac {\sin \left( 2\,r \right) }{\sqrt {r+x}}}\,{\rm d}r,\qquad x>0. $$ The proof that $S$ is not elementary may also work for this?
Part I. The following two formulas \begin{equation} {}_1F_2\biggl(1; n+1, n+\frac{3}{2}; -\frac{x^2}{4}\biggr) =(-1)^n\frac{(2n+1)!}{x^{2n+1}} \Biggl[\sin x-\sum_{k=0}^{n-1} (-1)^k\frac{x^{2k+1}}{(2k+1)!}\Biggr] \end{equation} and \begin{equation} {}_1F_2\biggl(1;n+\frac{1}{2}, n+1; -\frac{x^2}{4}\biggr) =(-1)^n\frac{(2n)!}{x^{2n}} \Biggl[\cos x-\sum_{k=0}^{n-1} (-1)^k\frac{x^{2k}}{(2k)!}\Biggr] \end{equation} for $n\in\mathbb{N}$ and $x>0$ were published on Page 16 of the reference [1] below. On the other hand, we know \begin{equation} {}_1F_2\biggl(1;1,\frac{3}{2}; -\frac{x^2}{4}\biggr) =\frac{4 \operatorname{arcsinh}\bigl(\frac{x}{2}\bigr)}{x \sqrt{x^2+4}\,} \end{equation} and \begin{equation} {}_1F_2\biggl(1;\frac{1}{2},1; -\frac{x^2}{4}\biggr) =\frac{2}{\sqrt{x^2+4}\,}. \end{equation} These four formulas reveal that the hypergeometric functions $$ {}_1F_2\biggl(1;n,n+\frac12;-x^2\biggr)\quad\text{and}\quad {}_1F_2\biggl(1;n-\frac12,n;-x^2\biggr) $$ for $n\in\mathbb{N}$ and $x>0$ are elementary! In other words, the hypergeometric function $$ {}_1F_2\biggl(1;\frac{n}2,\frac{n+1}2;-x^2\biggr), \quad n\in\mathbb{N} $$ has a closed-form expression.
Part II. Among other things, combining the above first two formulas with Theorems 1 and 2 in the reference [1] below, we can conclude that both of the hypergeometric function ${}_1F_2\bigl(1; n+1, n+\frac{3}{2}; -\frac{x^2}{4}\bigr)$ for $n\ge1$ and the hypergeometric function ${}_1F_2\bigl(1;n+\frac{1}{2}, n+1; -\frac{x^2}{4}\bigr)$ for $n\ge2$ are positive and decreasing in $x\in(0,\infty)$, while both of them are concave in $x\in(0,\pi)$. Summing up, the hypergeometric function $$ {}_1F_2\biggl(1; \frac{n+3}{2},\frac{n+4}{2}; -\frac{x^2}{4}\biggr), \quad n\in\mathbb{N} $$ is positive and decreasing in $x\in(0,\infty)$, while it is concave in $x\in(0,\pi)$.
Part III. Considering the first two formulas in Part I above, we observe that, in the reference [2] below, among other things, the function \begin{equation*} \ln\biggl[{}_1F_2\biggl(1;n+\frac{1}{2}, n+1; -\frac{x^2}{4}\biggr)\biggr], \quad n\in\mathbb{N} \end{equation*} was expanded into a Maclaurin power series at the point $x=0$, as well as the function \begin{equation*} \frac{\ln\bigl[{}_1F_2\bigl(1;\frac{5}{2}, 3; -\frac{x^2}{4}\bigr)\bigr]}{\ln\cos x} \end{equation*} was proved to be decreasing on $\bigl(0,\frac{\pi}2\bigr)$.
For more information, please refer to Remark 7 in the paper [3] below.
References
Your question posed two very hard problems.
If $a$ is continuous, the generalized hypergeometric stays not simplified and an interpretation with respect to elementary function is not possible.
From the use of indices, You gave two groups of values:
$\{1\}$ for the first lower index $1$
and
$\{a,a+\frac{1}{2}\}$ for the second lower index $2$ behind the symbol $F$.
I need a workaround to the condition $x<0$. For this is use $-abs(x)$.
So if I calculate
$_1F_2(\{1\};\{a,a+\frac{1}{2}\};-abs(x))$
for $a=2$, I get
$\frac{3\left(2\sqrt{|x|}-\sin\left(2\sqrt{|x|}\right)\right)}{4|x|^{3/2}}$
This is a composition of elementary functions. But it is certainly not an elementary function. For noninteger $a$ no such closed form is given.
For $a=3$, I get
$\frac{5\left(4|x|^{3/2}-6\sqrt{|x|}+3\sin\left(2\sqrt{|x|}\right)\right)}{4|x|^{5/2}}$
For $a=4$, I get
$\frac{5\left(4|x|^{3/2}-6\sqrt{|x|}+3\sin\left(2\sqrt{|x|}\right)\right)}{4|x|^{5/2}}$.
So the trend is clear more and more $\frac{1}{2}$ potences appear in the polynomial. For real $a$ the solution are continuous in between the integer ones.
The negative definition range alters the $\sinh$ to the $\sin$ that is all.
For $a=1$ the simplification is $\frac{\sin\left(2\sqrt{|x|}\right)}{2\sqrt{|x|}}$. So this is more elementary than the sums. The composition is only the quotient of two elementary functions.
Might be useful - an expression in terms of the incomplete Gamma function. The series $$ {}_1F_2(1;a,a+\frac12;-x^2)=1-\frac{4x^2}{2a(2a+1)}+\frac{16x^4}{2a(2a+1)(2a+2)(2a+3)}-\frac{64x^6}{2a(2a+1)(2a+2)(2a+3)(2a+4)(2a+5)}+... $$ can be written as $$ \frac12\left(f(2ix)+f(-2ix)\right), $$ where $$ f(t)=1+\frac t{2a}+\frac{t^2}{2a(2a+1)}+\frac{t^3}{2a(2a+1)(2a+2)}+...=\frac{2a-1}{t^{2a-1}}e^t(\Gamma(2a-1)-\Gamma(2a-1,t)) $$
