Griffiths Quantum Mechanics 2nd Edition: Problem 1.15

This is a pretty fun problem. It asks us to consider an unstable probability, that will spontaneously decay at some given time, and as such, the probability of finding the particle anywhere will decrease with time. We can define the probability of finding this particle anywhere at time $t$ as:

$$P(t) \ = \ \displaystyle\int_{-\infty}^{\infty} \ |\Psi(x, \ t)|^2 \ \text{dx} \ = \ e^{-t/T}$$
Where $T$ is defined as the lifetime of the particle (how long it takes the probability of the particle being found somewhere to decay). Evidently, if $T$ is very large, the particle's probability will decay at a slower rate, suggesting a more stable particle, while if $T$ is small, the particle will have a high probability of decaying much faster. Now, the problem asks us to consider the case where $V$ is not real. Up until now, when we have taken the complex conjugate of the Schrödinger equation, we have placed our signs under the assumption that $V$ is real. Now, we assign $V$ an imaginary part:

$$V \ = \ V_0 \ - \ i\Gamma$$
Where $V_0$ is the actual potential energy function, and $-i\Gamma$ is the imaginary component, with $\Gamma$ being some real constant. We now must show that with this new definition of $V$, that:

$$\frac{\text{d}P}{\text{dt}} \ = \ -\frac{2\Gamma}{\hbar} P$$
This proof is fairly straightforward. Let us start with the definition of $P$, then take the time derivative and expand:

$$P(t) \ = \ \displaystyle\int_{-\infty}^{\infty} \ |\Psi(x, \ t)|^2 \ \text{dx} \ \Rightarrow \ \frac{\text{d}P}{\text{dt}} \ = \ \frac{\text{d}}{\text{dt}} \ \displaystyle\int_{-\infty}^{\infty} \ |\Psi(x, \ t)|^2 \ \text{dx} \ = \ \displaystyle\int_{-\infty}^{\infty} \frac{\partial}{\partial t} \ |\Psi(x, \ t)|^2 \ \text{dx}$$ $$\Rightarrow \ \displaystyle\int_{-\infty}^{\infty} \frac{\partial}{\partial t} \ |\Psi(x, \ t)|^2 \ \text{dx} \ = \ \displaystyle\int_{-\infty}^{\infty} \Psi^{*}\frac{\partial \Psi}{\partial t} \ + \ \Psi\frac{\partial \Psi^{*}}{\partial t} \text{dx}$$ $$\Rightarrow \ \displaystyle\int_{-\infty}^{\infty} \Psi^{*}\Big( \frac{i\hbar}{2m} \frac{\partial^2 \Psi}{\partial x^2} \ - \ (\frac{i}{\hbar}V_0 \ + \ \frac{\Gamma}{\hbar})\Psi \Big) \ + \ \Psi\Big( -\frac{i\hbar}{2m} \frac{\partial^2 \Psi^{*}}{\partial x^2} \ + \ (\frac{i}{\hbar}V_0 \ - \ \frac{\Gamma}{\hbar})\Psi^{*} \Big) \text{dx}$$ $$\Rightarrow \ \displaystyle\int_{-\infty}^{\infty} \Psi^{*}\Big( \frac{i\hbar}{2m} \frac{\partial^2 \Psi}{\partial x^2} \ - \ \frac{\Gamma}{\hbar}\Psi \Big) \ + \ \Psi\Big( -\frac{i\hbar}{2m} \frac{\partial^2 \Psi^{*}}{\partial x^2} \ - \ \frac{\Gamma}{\hbar}\Psi^{*} \Big) \text{dx}$$ $$\Rightarrow \ \frac{i\hbar}{2m} \ \displaystyle\int_{-\infty}^{\infty} \ \Psi^{*}\frac{\partial^2 \Psi}{\partial x^2} \ - \ \Psi\frac{\partial^2 \Psi^{*}}{\partial x^2} \text{dx} \ -\frac{2\Gamma}{\hbar} \ \displaystyle\int_{-\infty}^{\infty} \ \Psi^{*} \ \Psi \text{dx} \ = \ \frac{i\hbar}{2m} \ \displaystyle\int_{-\infty}^{\infty} \ \frac{\partial}{\partial x} \Big( \Psi^{*}\frac{\partial \Psi}{\partial x} \ - \ \Psi\frac{\partial \Psi^{*}}{\partial x} \Big) \text{dx} \ -\frac{2\Gamma}{\hbar} \ P$$ $$\Rightarrow \ \frac{i\hbar}{2m} \ \Big( \Psi^{*}\frac{\partial \Psi}{\partial x} \ - \ \Psi\frac{\partial \Psi^{*}}{\partial x} \Big)\biggr\rvert_{-\infty}^{\infty} \ - \ \frac{2\Gamma}{\hbar} \ P \ = \ - \frac{2\Gamma}{\hbar} \ P$$

Ok, so we have proved the first statement. Now comes the task of finding the lifetime of the given particle in terms of $\Gamma$. We can set up this equation:
$$\frac{\text{d}}{\text{d}t} e^{-t/T} \ = \ -\frac{2\Gamma}{\hbar} e^{-t/T}$$

Now, this is a fairly simple differential equation that can be solved, however, I solved this equation by simply by making the observation that after differentiating, $e^{-t/T}$ cancels out, leaving us to solve only for $T$:
$$\frac{\text{d}}{\text{d}t} e^{-t/T} \ = \ -\frac{1}{T} e^{-t/T} \ \Rightarrow \ -\frac{2\Gamma}{\hbar} e^{-t/T} \ = \ -\frac{1}{T} e^{-t/T} \ \Rightarrow \ -\frac{2\Gamma}{\hbar} \ = \ -\frac{1}{T} \ \Rightarrow \ T \ = \ \frac{\hbar}{2\Gamma}$$