I haven't actually done the computation, but it seems to me that integrating the $\Phi(x)$ term by parts ad nauseam, you get a nice power series for $h(x).$
EDIT @Davide's argument is obviously the complete answer to the question as asked, but just as a coda, the series for $h(x)$ is quite cute:
In the odd part, the coefficients of $x^{2k+1}$ is $1/p(k)$ where $p(k)$ is the product of the first $k$ odd integers, while in the even part, the coefficient of $x^{2k}$ is $\sqrt{\pi/2}/q(k),$ where $q(k)$ is the product of the first $k$ even integers.
