I Why Does Unstable Particle Decay Follow an Exponential Curve?

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Given that an unstable particle has a constant probability of decaying per unit time, why is it said that its chance of surviving falls exponentially?
 
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It's like froth on beer: the more there is, the more there decays.
Or (in the other direction) like money accumulating on a bank: the more there is, the more interest (in absolute sense) you get, which gains interest again, etc.

Probability ##p## is given, number of decays is proportional to number of particles ##N##, so $$ {dN\over dt} = - p N $$.

At an "I" level you can solve this kind of differential equation
 
It has a constant probability to decay per unit time if it still lives. This survival probability goes down over time, therefore the probability to see a decay after time x goes down with increasing x. If you solve the differential equation, you see that it goes down exponentially.
 
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I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

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