This one weird infinite product can define exponentials in terms of itself. What does it do for other constants? What is... $$\lim_{\omega \to \infty} 
\prod_{N=1}^{\omega} {{1+e^{b \cdot c^{-N}}} \over 2}$$
This is similar to my other question. However, there is a constant factor rather than variable in the product.
My attempt: I have absolutely no clue except for the case of $c=2$ and  $b=b$
Create a line integral over the unit line evaluated with a uniform measure...
$$\int_L e^{b \cdot x} d \mu=\int_{L/2} e^{b \cdot x} \ d\mu+\int_{L/2} e^{b \cdot (x+1/2)} \ d\mu$$
This identity should be evident by self-similarity. Prepare for recursion...
$$\int_L e^{b \cdot x} d \mu=\int_{L/2} e^{b \cdot x}+e^{b \cdot (x+1/2)} \ d\mu=(1+e^{b/2}) \cdot \int_{L/2} e^{b \cdot x} \ d\mu$$
$$\Rightarrow \int_L e^{b \cdot x} d \mu=(1+e^{b/2}) \cdot \left( \int_{L/4} e^{b \cdot x} \ d\mu+\int_{L/4} e^{b \cdot (x+1/4)} \ d\mu \right)$$
$$\Rightarrow \int_L e^x d \mu=(1+e^{b/2}) \cdot (1+e^{b/4}) \cdot \left( \int_{L/4} e^{b \cdot x} \ d\mu \right)$$
It wouldn't be hard to prove by induction then that...
$$\Rightarrow \int_L e^{b \cdot x} d \mu=\lim_{\omega \to \infty} 
\prod_{N=1}^{\omega} (1+e^{b \cdot 2^{-N}}) \cdot \int_{L/{2^{\omega}}} e^{b \cdot x} \ d\mu$$
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 $0$...
$${{e^b-1} \over b}= \lim_{\omega \to \infty} 
\prod_{N=1}^{\omega} {{{1+e^{2^{-N}}}} \over 2}$$
Numerical analysis
For $b=2$, the product for $c=2$ is $3.1045...$, $c=3$ is $1.7532..$, $c=4$ is $1.4424$ and $c=5$ is $1.3108$. Plotting seems to suggest an exponential in the denominator.
Motivation: Getting an answer will allow me to derive methods to integrate a function like $e^x$ over fractals. If you doubt this, look here.
Specifically,
$$\int_C e^{b \cdot x} \ d \mu=\lim_{\omega \to \infty} 
\prod_{N=1}^{\omega} {{1+e^{2 \cdot b \cdot {3}^{-N}}} \over 2}$$
Where the region of integration $C$ is the cantor set
 A: The case $c=2$ is equivalent to the classical formula
$$ \sin(x) = x \prod_{k=1}^\infty \cos(x/2^k)$$
and its hyperbolic form
$$ \sinh(x) = x \prod_{k=1}^\infty \cosh(x/2^k)$$
I'm not aware of anything similar for $c \ne 2$, however.  
If $P(x) = \prod_{k=1}^\infty \cos(x c^{-k})$, we have  the functional equation
$P(cx) = \cos(x) P(x)$.  For $c=2$, $P(x) = \sin(x)/x$ is a solution to this.  I don't know if there's a closed-form
solution for $c \ne 2$. 
EDIT: In the correspondence with your formula, $x = b/2$.  Note that 
$$ \dfrac{1 + e^{b c^{-k}}}{2} =  e^{b c^{-k}/2} \cosh(b c^{-k}/2)$$ 
Now for $c > 1$,
$$ \prod_{k=1}^\infty e^{b c^{-k}/2}  = \exp\left(\sum_{k=1}^\infty b c^{-k}/2\right) = \exp\left(\frac{b}{2c-2}\right)$$
so
$$ \prod_{k=1}^\infty \dfrac{1 + e^{b c^{-k}}}{2} = \exp\left(\frac{b}{2c-2}\right) \prod_{k=1}^\infty \cosh(bc^{-k}/2) $$
EDIT: To prove the "classical formula", you can start with
$$ \sin(x/2^n) \prod_{k=1}^n \cos(x/2^k) = \sin(x)/2^n$$
which is easy to prove using induction.  Divide by $\sin(x/2^n)$ and take the limit as $n \to \infty$.
