= Solution
The proposed <probability density function> is nonnegative. To check its integral, put $I=\int_{\mathbb R}e^{-x^2/2}\,dx$. The <Gaussian integral> can be evaluated by squaring and using polar coordinates:
$$
I^2=\int_{\mathbb R^2}e^{-(x^2+y^2)/2}\,dx\,dy=2\pi\int_0^\infty re^{-r^2/2}\,dr=2\pi.
$$
Since $I>0$, $I=\sqrt{2\pi}$ and $\int\phi=1$. Thus $X$ has the <standard normal distribution>. Completing the square in its <moment-generating function> gives, for every real $t$,
$$
M_X(t)=\frac1{\sqrt{2\pi}}\int_{\mathbb R}e^{tx-x^2/2}\,dx=e^{t^2/2}\frac1{\sqrt{2\pi}}\int_{\mathbb R}e^{-(x-t)^2/2}\,dx=\boxed{e^{t^2/2}}.
$$
Differentiation under the integral is justified near every finite $t$ by Gaussian decay. Expanding $e^{t^2/2}=\sum_{m\geq0}t^{2m}/(2^m m!)$ gives \b[all moments]:
$$
\boxed{\mathbb E[X^{2m+1}]=0,\qquad\mathbb E[X^{2m}]=\frac{(2m)!}{2^m m!}\quad(m\geq0).}
$$
To obtain the <two-term bounds for Mills ratio>, use $\phi'(t)=-t\phi(t)$. <Integration by parts> gives, for $x>0$,
$$
\int_x^\infty\phi(t)\,dt=\frac{\phi(x)}x-\int_x^\infty\frac{\phi(t)}{t^2}\,dt.
$$
The positive remainder proves $r(x)<1/x$. A second <integration by parts> gives
$$
\int_x^\infty\frac{\phi(t)}{t^2}\,dt=\frac{\phi(x)}{x^3}-3\int_x^\infty\frac{\phi(t)}{t^4}\,dt.
$$
Substitution and division by the strictly positive $\phi(x)$ show
$$
r(x)=\frac1x-\frac1{x^3}+\frac3{\phi(x)}\int_x^\infty\frac{\phi(t)}{t^4}\,dt.
$$
The last integral is strictly positive for every finite $x>0$, proving \b[$\boxed{1/x-1/x^3<r(x)<1/x}$]. Both remainders are finite for every fixed positive $x$; the lower bound remains valid even when it is negative.
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