When the decay constant is not constant. Limit definition of the exponential of an integral? In radioactive decay (for example) the probability for a particle to decay per unit time is $\Gamma$. When this is a constant the probability to not decay after time $T$, $P(t)$, is derivable by splitting $T$ into $n$ timesteps of length $dt = T/n$ and using
$$P(t) = \lim_{n \rightarrow \infty}\left( 1 - \Gamma \frac{T}{n}\right)^n = e^{-\Gamma T},$$
using the limit definition of the exponential.
I am interested in the case when $\Gamma \rightarrow \Gamma(t)$ is a function of time. I suspect the answer is now
$$P(t) = e^{-\int_0^T \Gamma(t) dt}.$$
I am wondering if one can prove this again by the limit definition of an exponential. Discretising time into $n$ steps again I believe
$$P(t) = \lim_{n\rightarrow \infty}\Pi_{k=0}^n \left( 1 - \Gamma(k T/n) \frac{T}{n}\right).$$
Is this known to be an alternative definition of $e^{-\int_0^T \Gamma(t) dt}$?
Is there a better way to prove this?
 A: This is the first thing that came to mind, it may not be very well controlled. Take logarithms and get (setting $\Delta t=T/n$ for legibility):
$$\log P = \lim_{n\to \infty} \sum_{k=1}^{k=n}\log (1-\Gamma(k\Delta t)\Delta t)$$
As $n\to \infty$ we have $\Delta t\to 0$ thus, using $\log(1+x)=x+\mathcal{O}(x^2)$:
$$\log P = -\lim_{n \to \infty} \sum_{k=1}^{k=n} \Gamma(k\Delta t)\Delta t + \mathcal{O}(n\Delta t^2)$$
the first term is the Riemann sum definition of the integral you want. The second term should generally tend to zero but there may be some pathological functions $\Gamma$ that prevent this. In any case, the equation OP suggets is almost certainly okay in physically relevant situations.
A: If you consider a small time interval $dt$, the change of probability to not decay ($dP(t)$) is given by the product of probability to decay per unit time ($\Gamma$) times time interval ($dt$) times the current probability that the particle has not decayed yet ($P(t)$):
$$dP(t) = -\Gamma(t) dt P(t).$$
So, basically we derived a differential equation on $P(t)$. Your proposed
$$P(T) = e^{-\int_0^T \Gamma(t)dt}$$
is indeed a solution to this equation. Moreover, it obeys the correct initial condition:
$$P(T=0) = 1$$
which means the particle begins to decay at the moment $T=0$.
This logic could be used to prove your relation for the exponential of an integral as well, even though we don't need to use it in this approach.
