Finding the sum- $x+x^{2}+x^{4}+x^{8}+x^{16}\cdots$ 
If $S = x+x^{2}+x^{4}+x^{8}+x^{16}\cdots$
Find S.

Note:This is not a GP series.The powers are in GP.
My Attempts so far:
1)If $S(x)=x+x^{2}+x^{4}+x^{8}+x^{16}\cdots$
Then $$S(x)-S(x^{2})=x$$
2)I tried finding $S^{2}$ and higher powers of S to find some kind of recursive relation.
3)When all failed I even tried differentiating and integrating S.Obviously that was of no good either.
Could anyone give me a hint to solve this?Thanks! 
 A: I have worked on this series before.
There is no simple closed form as a geometric series has.
(such as $x+x^2+x^3+x^4+...=\frac{x}{1-x}$ where $|x|<1$).
You can see more information in the link about Lacunary function.
The series can be expressed in closed form of double integral. I shared my result below.
$x+x^{2}+x^{4}+x^{8}+\dots=F(x)$ 
Let's transform 
$x=e^{-2^t} \tag{1}$  
Where $-\infty<t<\infty$
Thus x will be in $ (0,1)$ and  $F(x)$ is not divergent in this range.
$e^{-2^t}+e^{-2^{t+1}}+e^{-2^{t+2}}+\dots=F(e^{-2^t})=H(t)$
$e^{-2^t}+H(t+1)=H(t)$
$H(t+1)-H(t)=-e^{-2^t}$

The Fourier transform of both sides  
$$\int_{-\infty}^{+\infty} H(t+1)e^{-2πift} \mathrm{d}t-\int_{-\infty}^{+\infty} H(t)e^{-2πift} \mathrm{d}t=-\int_{-\infty}^{+\infty} e^{-2^{t}}e^{-2πift} \mathrm{d}t$$
$$V(f)= \int_{-\infty}^{+\infty} H(t)e^{-2πift} \mathrm{d}t$$
$$\int_{-\infty}^{+\infty} H(t+1)e^{-2πift} \mathrm{d}t=\int_{-\infty}^{+\infty} H(z)e^{-2πif(z-1)} \mathrm{d}z=V(f)e^{2πif}$$
$$e^{2πif}V(f)-V(f)=-\int_{-\infty}^{+\infty} e^{-2^{t}}e^{-2πift} \mathrm{d}t$$
$$V(f)=\int_{-\infty}^{+\infty} \frac{e^{-2^{t}}e^{-2πift}}{1-e^{2πif}} \mathrm{d}t$$ 

Now we need to take the inverse Fourier transform  
$$H(z)=\int_{-\infty}^{+\infty} V(f) e^{2πifz} \mathrm{d}f=\int_{-\infty}^{+\infty}  e^{2πifz}\int_{-\infty}^{+\infty} \frac{e^{-2^{t}}e^{-2πift}}{1-e^{2πif}} \mathrm{d}t\,\mathrm{d}f $$
The closed form of $H(z)$ in integral expression:
$$H(z)=\int_{-\infty}^{+\infty}  \int_{-\infty}^{+\infty} e^{2πifz} \frac{e^{-2^{t}}e^{-2πift}}{1-e^{2πif}}  \mathrm{d}t\,\mathrm{d}f $$ 
$$\sum_{k=0}^\infty x^{2^k}=H(\log_2(-\ln x))=\int_{-\infty}^{+\infty}  \int_{-\infty}^{+\infty} e^{2πif\log_2(-\ln x)} \frac{e^{-2^{t}-2πift}}{1-e^{2πif}}  \mathrm{d}t\,\mathrm{d}f$$ 
Where $0<x<1$
$$\sum_{k=0}^\infty x^{2^k}= \int_{-\infty}^{+\infty} \frac{e^{2πif\log_2(-\ln x)}}{1-e^{2πif}} \int_{-\infty}^{+\infty} e^{-2^{t}-2πift}  \mathrm{d}t\,\mathrm{d}f=\int_{-\infty}^{+\infty}  e^{-2^{t}} \int_{-\infty}^{+\infty}  \frac{e^{2πif(\log_2(-\ln x)-t)}}{1-e^{2πif}}  \mathrm{d}f\,\mathrm{d}t$$
Note: I made the update on 07/29/2016 . Variable change was $x=e^{2^t}$ at Tag (1) and it had problems as @leonbloy 's comment below. Thanks for the comment. Please let me know if you notice something else in definitions.
