= Solution
For the trial function, $y_T'=-ax\,y_T$. The required <Gaussian integrals> are
$$
G[y_T]=\int_{\mathbb R}e^{-ax^2}\,dx=\sqrt{\frac\pi a},\qquad\int_{\mathbb R}x^2e^{-ax^2}\,dx=\frac{\sqrt\pi}{2a^{3/2}}.
$$
The second formula follows by differentiating the first with respect to $a$. Therefore
$$
F[y_T]=a^2\frac{\sqrt\pi}{2a^{3/2}}-V_0\sqrt{\frac\pi{1+a}},
$$
and division by $G[y_T]$ gives the <Gaussian variational bound for an attractive Gaussian well>:
$$
\boxed{E(a)=\frac a2-V_0\sqrt{\frac a{1+a}},\qquad a>0.}
$$
This is a good trial family because it consists of smooth, positive, <square-integrable functions> decaying at both infinities. Each is an <even function>, matching the symmetry of the lowest <bound state> in this even potential, and the parameter varies its width. It also permits exact <Gaussian integrals>. Its Gaussian tail need not equal the exact exponential tail of a negative-energy <bound state> for the <Rayleigh-Ritz variational principle> to give a valid bound.
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