integral of a discontinous periodic function?

Question

Background

I am a biology student and I have been struggling with finding the integral of the function:

$$ a(t) = a_0\times e^{-k\times(t- \lfloor t/\tau \rfloor * \tau)} \space\space\space \text{with}\space (k \neq 0, \tau \neq0)$$

Which (for some arbitrary values for $a_0, k$, $\tau $) looks like this. A fancy version of exponential decay.

My understanding of this function is

1. periodic (for a $\tau > 0$), therefore I only need to determine the integral over one period.
2. non-continuous /discontinuous.

What I did: I found the integral of the "non-periodic" version of $a(t)$, lets call that one $\alpha(t)= a_0*e^{-kt}$. So,$$\int \alpha(t) dt = -\frac{a_0*e^{-kt}}k $$ I then simply did $$\int_0^{\tau} \alpha(t) dt = -\frac{a_0*e^{-k \times \tau}}k +\frac{a_0}k $$ but this way I am clearly missing that originally $a(\tau)= a_0$ (depicted by the black points in the image above).

My Question

Can I use $\alpha(t)$ instead of $a(t)$ as to integrate over the interval fomr 0 to $\tau$? I am asking because obviously $a(tau)\neq\alpha(tau)$, but I really can't think of another way.

Bigger picture: $a(t)$ is a simplified model of the concentration of a "pollutant" in a waterbody. The concentration of this "pollutant" influences the growth of microorganisms. By finding $ \int_0^{\tau} a(t) $ I ultimately want to determine the respective mean growth of microorganisms over $\tau$. But one step at a time. I want to do this analytically, not numerically.

Answer 1

For $0 \leqslant t \leqslant \tau$, the term $\lfloor t/\tau \rfloor = 0$. Then your integral is correct save only you have dropped the factor $a_0$. But if you want to integrate over period greater than $\tau$, you need to consider the integral over the initial period: you will end up with a sum for a whole number of $\tau$ length periods and a residual.

For instance when $T \geqslant \tau$, let $\lfloor T/\tau \rfloor = n$, \begin{align} \int_0^T \alpha(t)~ dt &= a_0\int_0^{\tau} e^{-kt} dt + a_0\int_{\tau}^{2\tau} e^{-k(t-\tau)}~dt + \cdots \\ &\quad\qquad + a_0\int_{(n-1)\tau}^{n\tau} e^{-k(t-(n-1)\tau)}~dt+ a_0\int_{n\tau}^T e^{-k(t-n\tau)}~dt \\ &=a_0\int_0^{\tau} e^{-kt} dt + a_0\int_{0}^{\tau} e^{-kt}~dt + \cdots \\ &\quad\qquad + a_0\int_{0}^{\tau} e^{-kt}~dt+ a_0\int_{0}^{T-n\tau} e^{-kt}~dt \\ &=a_0\frac{\lfloor T/\tau\rfloor}{k}(1-e^{-k\tau})+\frac{a_0}{k}(1-e^{-k(T-\lfloor T/\tau\rfloor \tau)}). \end{align} I hope this is helpful.

Answer 2

WLOG, $a_0=1$ and $\tau=1$. The integral is

$$\int_0^te^{-k\{u\}}du$$ (the braces denote the fractional part). Within the first period ($t<1$),

$$\int_0^te^{-ku}du=\frac{1-{e^{-kt}}}k.$$ For a complete period,

$$\int_0^1e^{-ku}du=\frac{1-{e^{-k}}}k.$$

Then for any value,

$$\int_0^t=\int_0^{\lfloor t\rfloor}+\int_{\lfloor t\rfloor}^t=\lfloor t\rfloor\int_0^1+\int_0^{\{t\}}.$$

You don't have to worry about the discontinuity $f(1)\ne f(0)$. It has null measure and does not influence the integral.

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Below, $k=1$: