= Solution
Complete the square in the integral against the <standard normal density>:
$$
\begin{aligned}
M_X(t)&=\frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{tx-x^2/2}\,dx\\
&=e^{t^2/2}\frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-(x-t)^2/2}\,dx
=\boxed{e^{t^2/2}.}
\end{aligned}
$$
The remaining integral is one because its integrand is the <probability density function> of the <normal distribution> $N(t,1)$. Hence the <moment-generating function of a standard normal variable> is finite on the whole real line.
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