$$\sum_{k=0}^{\infty}(1-e^{-x/2^k}) \space \Rightarrow \space \frac{\Gamma(s)}{2^s-1},\space(-1<\text{Re}(s)<0)$$ by Mellin Transform, and its asymptotics (as $\space x\to\infty$) by Converse mapping $$ \log_{2}{x} + \frac{\gamma}{\log2} + \frac{1}{2}+P(x)+O(x^{-1})$$ where P() is a $\Gamma()$ based periodic func(not an issue here)
I want to know how to get the constant terms, $\frac{\gamma}{\log2} + \frac{1}{2}$.
There are poles at s = 0(double), pure imaginary(->P(x)) and non-negative integers(don't care, because $\space x\to\infty$, Only Right-sided poles considered)
I got $\log_{2}{x}$ from $\lim_{s\to0} s^2F(s)$, and got $\frac{\gamma}{\log2}$ from $\lim_{s\to0} sF(s) = \frac{\Gamma(s+1)}{2^s-1} \to \frac{\Gamma'(s+1)}{2^s\log2} = \frac{-\gamma}{\log2}$
Then, Where the hell does $\frac{1}{2}$ come from? (shouldn't the constants come only from order $1$ & pole $s = 0$?)
when you consider $$2\pi if(x)=\int_{c=\Re s}\frac{\Gamma(s)}{2^s-1}x^{-s}ds, -1 < c <0$$ you notice that the integrand goes very fast to zero on any vertical line as $|t| \to \infty$ (except on $\Re s =0$ where $2^s-1$ has zeroes going to infinity too), uniformly for $a \le \Re s \le b$ as long as you stay away from the zeroes of $2^s-1$, so you can use Cauchy on large rectangles with the horizontal part passing through $\pi \log 2 +2\pi k/log 2$ say so $|2^s-1| >> k>0$ there for all $a \le Re s \le b$ to pass to the integral on any line to the right and get
$$f(x)=\frac{1}{2\pi i}\int_{c=\Re s}\frac{\Gamma(s)}{2^s-1}x^{-s}ds- g(x), c >0$$ arbitrary, where $g(x)$ is given by the residues of the integrand on $\Re s =0$ since the integrand is analytic to the right of $0$
(the minus sign for the residues coming from the fact that the counterclockwise orientation tells us that $\int_{\Re s=b}f(s)ds-\int_{\Re s=a}f(s)=2\pi i$ sum residues in between the vertical lines at $a$ and $b$ when $b>a$)
The residues at $2\pi k/ \log 2, k \ne 0$ are handled as you mentioned, but at $s=0$ we have a double pole so the residue formula requires to take the derivative of $\frac{s^2\Gamma(s)}{2^s-1}x^{-s}$ and evaluate it at zero
(for a double pole at zero one has $F(s)=a_2/s^2+ a_1/s+G(s)$ so $s^2F(s)=a_2+a_1s +s^2G(s)$ so $a_1=(s^2F(s))'|_{s=0}$)
Here using $s\Gamma (s)=\Gamma(s+1)$ you get that the residue is the derivative of $\frac{s\Gamma(s+1)}{2^s-1}x^{-s}$ at zero and that derivative has three terms by the usual product rule where we take $(\frac{s}{2^s-1})'$ as one multiplier, the other two coming from $(x^{-s})'$ and $\Gamma'(s+1)$
Since $\frac{s}{2^s-1} \to 1/\log 2, s \to 0$ and $\Gamma'(1)=-\gamma, (x^{-s})'|_{s=0}=-\log x$ those last two terms give us indeed residues that are $-\log x/\log 2, -\gamma/\log 2$, respectively (remembering that we subtract the residues when we go from left $-1<c<0$ to right $c>0$ )
The term coming from $(\frac{s}{2^s-1})'=\frac{2^s-1-s2^s \log 2}{(2^s-1)^2}$ converges to $\lim_{s \to 0}\frac{-s2^s (\log 2)^2}{2(2^s-1)\log 2}=-1/2$ and since $\Gamma(1)=1, x^{-0}=1$ we get that extra $1/2$ term mentioned