Originally Posted by
Chris L T521 Keep in mind that $\displaystyle \displaystyle\int_{-\infty}^{\infty}e^{-ax^2}\,dx=\sqrt{\frac{\pi}{a}}$. (*)
I take it you mean the original problem is $\displaystyle \displaystyle\int_{-\infty}^{\infty}\exp\left(-\left[\frac{1}{4a}+\frac{i\hbar t}{2m}\right]k^2+ikx\right)\,dk$, because the 2m term appears out of nowhere in the second step.
For the moment, let's focus on the exponent:
$\displaystyle -\left(\dfrac{1}{4a}+\dfrac{i\hbar t}{2m}\right)k^2+ixk$.
Rearranging, we get
$\displaystyle -\dfrac{m+2ai\hbar t}{4am}\left[k^2-\dfrac{4amix}{m+2ai\hbar t}k\right]$ (Verify)
Complete the square to get
$\displaystyle -\dfrac{m+2ai\hbar t}{4am}\left[k-\dfrac{2amix}{m+2ai\hbar t}\right]^2-\dfrac{amx^2}{m+2ai\hbar t}$ (Verify)
With this, we can now evaluate this integral instead:
$\displaystyle \displaystyle\int_{-\infty}^{\infty}\exp\left(-\frac{m+2ai\hbar t}{4am}\left[k-\frac{2amix}{m+2ai\hbar t}\right]^2-\frac{amx^2}{m+2ai\hbar t}\right)\,dk$ $\displaystyle \displaystyle=\exp\left(-\frac{amx^2}{m+2ai\hbar t}\right)\int_{-\infty}^{\infty}\exp\left(-\frac{m+2ai\hbar t}{4am}\left[k-\frac{2amix}{m+2ai\hbar t}\right]^2\right)\,dk$
Now apply the substitution $\displaystyle u=k-\dfrac{2amix}{m+2ai\hbar t}\implies \,du=\,dk$.
Thus, the integral becomes
$\displaystyle \displaystyle\exp\left(-\frac{amx^2}{m+2ai\hbar t}\right)\int_{-\infty}^{\infty}\exp\left(-\frac{m+2ai\hbar t}{4am}u^2\right)\,du$.
Now apply (*) to get
$\displaystyle \exp\left(-\dfrac{amx^2}{m+2ai\hbar t}\right)\sqrt{\dfrac{4am\pi}{m+2ai\hbar t}}=\dfrac{\exp\left(-\dfrac{ax^2}{1+2ai\hbar t/m}\right)}{\sqrt{\dfrac{m+2ai\hbar t}{4am\pi}}}=\dfrac{\exp\left(-\dfrac{ax^2}{1+2ai\hbar t/m}\right)}{\sqrt{\dfrac{1}{4a\pi}+\dfrac{i\hbar t}{2m\pi}}}$
which appears to be different from the answer you have supplied....so maybe I made a mistake somewhere??
But this is the idea of how to approach this problem. Hope this helps!