While computing the integral
$$\displaystyle\int_0^1{\displaystyle\sum_{n=1}^{\infty}x^{(2^n)}dx}$$
I easily got to
$$\displaystyle\sum_{n=1}^{\infty} \frac{1}{2^n+1}$$
Since this was getting ridiculously hard for me I resigned and asked WolframAlpha which gave me this solution, although without providing a procedure.
My confusion is unmeasurable. The second addend I believe we can all agree has to do with the antiderivative of $2^n$ but what perplexes me is the nominator, where did it come from and is the complex part of the approximation relevant or just a computer mistake? Any answer is accepted.
Although $\psi_{1/2}^{(0)}$ was passed a non-real argument $z:=1-\frac{i\pi}{\ln2}$, the result $w:=\psi_{1/2}^{(0)}(z)$ is real, as is the series $\frac{w}{\ln2}-1$. Since$$\Gamma_{1/2}(x)=\prod_{j\ge0}\frac{1-2^{-j-1}}{1-2^{-j-x}}2^{x-1}\implies\frac{\psi_{1/2}^{(0)}(x)}{\ln2}=\sum_{j\ge0}\frac{2^x-2^{1-j}}{2^x-2^{-j}},$$if $\Re x\in\frac{i\pi}{\ln2}\Bbb Z$ then $2^x=e^{x\ln2}\in\Bbb R$.