Answer
$\Gamma=0$
Work Step by Step
The answer to this question deals with proving that the separation constant $E$ must be real. This is done by contradiction by showing that if the separation constant were complex, the wavefunction will only be normalizable if and only if the imaginary part of the separation constant is zero.
$$
\Psi(x, t)=\psi(x) e^{-i E t / \hbar}
$$.
Normalisation means that:
$$
\int_{-\infty}^{+\infty}|\Psi(x, t)|^{2} d x=1
$$.
Set $E=E_0+i\Gamma$ and combine the two equations above.
$$
\int_{-\infty}^{+\infty}|\psi(x) e^{-i (E_0+i\Gamma) t / \hbar}|^{2} d x=\int_{-\infty}^{+\infty}|\psi(x)|^2 |e^{-i (E_0t)/ \hbar}|^2 |e^{ (\Gamma t)/ \hbar}|^2 dx= \int_{-\infty}^{+\infty}|\psi(x)|^2 \cdot 1 \cdot |e^{\Gamma t / \hbar}|^2 dx = |e^{\Gamma t}|^2 \cdot \int_{-\infty}^{+\infty}|\psi(x)|^2 dx
$$.
Since the normalisation condition must hold for all $t$, $\Gamma$ must be zero such that $|e^{\Gamma t}|^2=1$ and the wavefunction is normalised at all time.
