What is the limit of this divergent infinite product multiplied by an exponential? What is... $$\lim_{\omega \to \infty} 
\left( {1 \over {c^{\omega}}} \cdot  \prod_{N=1}^{\omega} (1+e^{b \cdot c^{-N}}) \right)$$
My attempt: I have absolutely no clue except for the case of $c=2$ and  $b=1$
Create a line integral over the unit line evaluated with a uniform measure...
$$\int_L e^x d \mu=\int_{L/2} e^x \ d\mu+\int_{L/2} e^{x+1/2} \ d\mu$$
This identity should be evident by self-similarity. Prepare for recursion...
$$\int_L e^x d \mu=\int_{L/2} e^x+e^{x+1/2} \ d\mu=(1+e^{1/2}) \cdot \int_{L/2} e^x \ d\mu$$
$$\Rightarrow \int_L e^x d \mu=(1+e^{1/2}) \cdot \left( \int_{L/4} e^x \ d\mu+\int_{L/4} e^{x+1/4} \ d\mu \right)$$
$$\Rightarrow \int_L e^x d \mu=(1+e^{1/2}) \cdot (1+e^{1/4}) \cdot \left( \int_{L/4} e^x \ d\mu \right)$$
It wouldn't be hard to prove by induction then that...
$$\Rightarrow \int_L e^x d \mu=\lim_{\omega \to \infty} 
\left(\prod_{N=1}^{\omega} (1+e^{2^{-N}}) \cdot \int_{L/{2^{\omega}}} e^x \ d\mu \right)$$
Yet we know what the left hand side equals, since it can be evaluated as a definite integral, also we know what the integral on the right equals. Since the measure is uniform and the number of values x will be allowed to take on the interval decreases to just the value, namely $0$...
$$e-1= \lim_{\omega \to \infty} 
\left( {1 \over {2^{\omega}}} \cdot \prod_{N=1}^{\omega} (1+e^{2^{-N}}) \right)$$
Motivation: Getting an answer will allow me to derive methods to integrate a function like $e^x$ over fractals.
 A: Define $y_k = \frac{1}{c^k} \prod_{i=1}^k(1 + e^{b/c^i})$.
Claim 1:  If $c>2$, then $\lim_{k\rightarrow\infty} y_k=0$.  If $1\leq c<2$, then $\lim_{k\rightarrow\infty} y_k = \infty$. 
Proof for case $c>2$:
Assume $c = 2 + \delta$ for some $\delta>0$.  There exists a value $k^*$ such that $e^{b/c^i} \leq 1 + \delta/2$ for all $i \geq k^*$. Then for all $k>k^*$ we have:
\begin{align} 
0 \leq y_k &= \frac{1}{c^k} \prod_{i=1}^k (1 + e^{b/c^i})\\
&= \left(\frac{1}{c^{k^*}}\prod_{i=1}^{k^*}(1+e^{b/c^i})\right)\frac{1}{c^{k-k^*}}\prod_{i=k^*+1}^{k} (1+e^{b/c^i})\\
&\leq \left(\frac{1}{c^{k^*}}\prod_{i=1}^{k^*}(1+e^{b/c^i})\right)\frac{(2+\delta/2)^{k-k^*}}{c^{k-k^*}}\\
&= \left(\frac{1}{c^{k^*}}\prod_{i=1}^{k^*}(1+e^{b/c^i})\right)\left(\frac{2+\delta/2}{2 + \delta}\right)^{k-k^*} \rightarrow 0\\
\end{align} 
So $\lim_{k\rightarrow\infty} y_k = 0$. 
Proof for case $1 < c < 2$: Similar.  
Proof for case $c=1$: $y_k = (1+e^{b/c})^k \rightarrow \infty$.

Claim 2: If $0 < c < 1$ and $b\geq 0$ then $\lim_{k\rightarrow\infty} y_k = \infty$. 
Proof: 
We know $1+e^{b/c^i} \geq 1$ for all $i \geq 1$, and so $y_k \geq 1/c^k\rightarrow\infty$.  

Claim 3: If $c=2$ and $b\geq 0$ then $1 \leq \lim_{k\rightarrow\infty} y_k \leq e^b$. 
Proof: Since $b \geq 0$ we know $1 \leq e^{b/2^i}$ for all $i$, and hence: 
\begin{align} 
y_k &= \frac{1}{2^k} \prod_{i=1}^k(1 + e^{b/2^i}) \\
&\leq \frac{1}{2^k} \prod_{i=1}^k (e^{b/2^i} + e^{b/2^i}) \\
&= \prod_{i=1}^k e^{b/2^i}
\end{align} 
Thus: 
\begin{align} 
\log(y_k) &\leq \sum_{i=1}^k \frac{b}{2^i} \rightarrow b
\end{align} 
and so $\lim_{k\rightarrow\infty} y_k \leq e^b$.

For an exact answer for $c=2, b\neq 0$, why not just apply your same method? 
\begin{align}
\int_0^1 e^{bx} dx &= \int_0^{1/2} e^{bx}dx + \int_0^{1/2} e^{b(x+1/2)}dx \\
&= (1+e^{b/2})\int_0^{1/2} e^{bx}dx \\
&= (1+e^{b/2})\left[ \int_0^{1/4} e^{bx}dx + \int_0^{1/4} e^{b(x+1/4)}dx\right]\\
&= (1+e^{b/2})(1+e^{b/4})\int_0^{1/4}e^{bx}dx
\end{align}
and so on, with the left-hand-side giving the answer of $\frac{e^b-1}{b}$? So if $c=2$ and $b\neq 0$ we get: 
$$ \lim_{k\rightarrow\infty} y_k = \frac{e^b-1}{b} $$
Notice that this is consistent with Claim 3, since $1 < \frac{e^b-1}{b} < e^b$ whenever $b>0$.
