I'd like to find the value of the following sum
$$S(u) = \sum_{n=0}^\infty \frac{e^{iu2^n}}{2^{n+1}}$$ for $u \in \mathbb R$, but I can't seem to do it.
Unfruitful work
Writing $$S = \sum_{n=0}^\infty \sum_{p=0}^\infty \frac{(iu2^n)^p}{2^{n+1}p!}$$ is proven useless because I can't exchange the order of summation.
Writing $$S = \sum_{n=0}^\infty \frac{\cos(u2^n)}{2^{n+1}} + i \sum_{n=0}^\infty \frac{\sin(u2^n)}{2^{n+1}}$$
seems also useless. Maybe it would be useful to find a fourier series that has those coefficients, but I have no idea on how to do that and anyhow since $u \in \mathbb R$ is not limited I don't think it can be done.
Someone has some ideas?
P.S.
If $X$ is a random variable with distribution $P(X=2^k) = \frac 1{2^{k+1}}$, then $S(u)$ is its charachteristic function. That is why I'm interested in finding it
There isn't a "closed form" solution in terms of standard functions. If you write $S(u) = F(z)/2$ with $z = e^{iu}$, $F(z) =\displaystyle \sum_{n=0}^\infty \dfrac{z^{2^n}}{2^n}$, then $ z F'(z) = \displaystyle \sum_{n=0}^\infty z^{2^n}$ is a commonly-used example of a lacunary function.