I have a problem with evaluating the following gaussian integral in two different ways \begin{equation} \psi(x,t) = \int_{\infty}^{\infty} e^{-\frac{\alpha^2 \hbar^2}{2} (k-k_0)^2} e^{(ikx - i\frac{\hbar k^2}{2m}t)}\; dk \end{equation} The most common approach is to make the substitution $u=(k-k_0)$. The integral now becomes \begin{equation} \psi(x,t) = \int_{\infty}^{\infty} e^{-\frac{\alpha^2 \hbar^2}{2} u^2} e^{(i(k_0+u)x - i\frac{\hbar (u^2 + 2uk_0 +k_0^2)}{2m}t)}\; du \end{equation} From this, we see that the linear term in $u$ has a time $(t)$ dependence. The result of this integral is a propagating wave part multiplied by a spreading gaussian: \begin{equation} \psi(x,t) = e^{(ik_0 x - i\frac{\hbar k_0 ^2}{2m}t)} e^{-\frac{(x-\frac{\hbar k_0}{m}t)^2}{2(\alpha^2 \hbar^2 + i\frac{2\hbar t}{m})}}. \end{equation}
Note: I ignored an overall factor because my problem is with the functional form. The problem is when I try to evaluate the integral without making the substitution. Then, there's no $t$ dependence in the linear term in $k$. I just took the formula for the gaussian integral with complex coefficients from the wiki page linked here. The integral now evaluates to \begin{equation} \psi(x,t) = e^{\frac{(\alpha^2 \hbar^2 k_0 + ix)^2}{-2(\alpha^2 \hbar^2 + \frac{i \hbar t}{m})}}, \end{equation} which clearly has a different $t$ dependence. But the answer shouldn't depend on whether or not I make the substitution, and I can't see where I went wrong.