How should I finish this by completing to square? I am trying to calculate the integral:
$$\int_{-\infty}^{+\infty}\exp\left[-\frac{a^2(k-k_0)^2}{2} + ikx-i\frac{hk^2}{2m}t\right]dk$$
I thought on maybe mark $k-k_0=u$ but I am not see how it gonna help me
P.S the integral is only with respect to $k$. $k_0$ is constant also like the rest of paramteters
$i$ is the imagenery unit

$$
-\frac{a^2(k-k_0)^2}{2} + ikx-i\frac{hk^2}{2m}t
= -\frac12 a^2 k^2 + a^2 k_0 k - \frac12 a^2 k_0^2 + ikx - i\frac12 \frac1m ht k^2 \\
= -\frac12 (a^2+i\frac{ht}{m}) k^2 + (a^2 k_0 + ix) k - \frac12 a^2 k_0^2 \\
= -\frac12 (a^2+i\frac{ht}{m}) \left( k^2 - 2 \frac{a^2 k_0 + ix}{a^2 + i\frac{ht}{m}} k + \frac{a^2 k_0^2}{a^2+i\frac{ht}{m}} \right) \\
= -\frac12 (a^2+i\frac{ht}{m}) \left( \left[ k - \frac{a^2 k_0 + ix}{a^2 + i\frac{ht}{m}} \right]^2 + \left[ \frac{a^2 k_0^2}{a^2+i\frac{ht}{m}} - \left(\frac{a^2 k_0 + ix}{a^2 + i\frac{ht}{m}}\right)^2 \right] \right) \\
$$
