I am trying to equate two integrals, one of exponential decay and another of exponential decay that drops with a step (floor) function. Specifically, I am trying to minimize the difference of these decays given values for $g_1$ and $s$ over $g_2$. Here is the resulting minimization:
$$ \operatorname*{argmin}_{g_2} \Bigg|\Bigg| \small \int_0^T g_1^{\lfloor t/s \rfloor} \, dt - \int_0^T g_2^t \, dt \Bigg|\Bigg| \\[10pt] \{g_1, g_2\in\mathbb{R} , 0 < g_1, g_2 < 1\} \\[10pt] \{T, S \in \mathbb{N}, 0 < s < T \} $$
We can assume that $T$ will be an integer multiple of $s$:
$$ \operatorname*{argmin}_{g_2} \Bigg|\Bigg| \small \int_0^{Ns} g_1^{\lfloor t/s \rfloor} \, dt - \int_0^{Ns} g_2^t \, dt \Bigg|\Bigg| \\[10pt] \{N \in \mathbb{N}, N>0\} $$
So far I am having no luck in find a solution to this.
I've tried converting the left-hand integral to a summation, and solving the right-hand integral, but I still get stuck at:
$$ s\sum_0^{Ns-1}{g_1^n} - \frac{g_2^{Ns} - 1}{\log g_2} $$
An approximate solution is also ok, but I didn't find the series expansion to help.
P.S. My math formatting may be wrong all over the place. If so, please help me correct it.
$$ \int_0^{Ns}g_1^{\lfloor t/s \rfloor} \, dt = \int_0^N g_1^{\lfloor u \rfloor} (s\,dt) = s \sum_{n=0}^{N-1} g_1^n = s \cdot \frac{g_1^N - 1}{g_1 - 1} = \frac{g_2^{Ns} - 1}{\log g_2} $$ So you have $$ \frac{g_2^{Ns} - 1}{\log g_2} = \text{constant} $$ where "constant" means not depending on $g_2$.
From $\displaystyle \frac{g_2^{Ns} - 1}{\log g_2} = c$ we get $\displaystyle g_2^{Ns} = 1+c\log g_2 = \log(g_2^c\cdot e).$
As $g_2\downarrow 1$, the fraction $\dfrac{g_2^{Ns} - 1}{\log g_2}$ approaches $Ns$. The fraction grows without bound as $g_2$ grows, so if $Ns<c$, then there must be some value of $g_2>1$ for which equality holds.
For this I think I would use Newton's method.