Evaluating $\mathrm{\int_0^1 ?(x)dx}$

Question

This function interestingly shown as ?(x) is dubbed the Minkowski Question Mark Function. It looks very similar to x. Wolfram Alpha can even plot the derivative of this apparently smooth function. Here are some details for what I want to find without using trivial solutions like the Riemann Sum definition nor $n^{th}$ derivatives in the answer from Taylor Series.$ \ a_k$ represents a continued fraction of x=$[a_1,…,a_k]$ as seen in the first top link.

$$q=\int_0^1 ?(x)dx=-2\int_0^1\limits \sum_k \frac{(-1)^k}{2^{\sum_{n=1}^ka_n}}dk≈.5$$

This brings up the property of $$?(x^{-1})=\frac1{2^{x-1}}\implies ?(x)=2^{1-\frac1x}, x=k^n\mathop \implies ^? q\mathop =^?\int_0^1 2^{1-\frac1x}dx=2\ln(2)\mathrm{Ei}(-\ln(2))+1=.475050468…≈.5$$

Here is what the constant looks amazingly like:

This result seems correct for a closed form with the exponential integral function:

It would be very nice to find a non-integral form of this constant which is not a trivial solution like the one described in paragraph one. This function does not looks smooth, but that does not mean we cannot integrate it. What is a better, closed form or not, alternative representation or form for the area under Minkowski’s Question Mark function as defined over the unit interval $x\in$[0,1]? Please correct me and give me feedback!

Result over half period because of self symmetry which seems to fail for a result:

$$\int_0^\frac12 ?(x)dx+\frac1{2^2}+1-\int_0^\frac12?(x)dx=\int_0^1 ?(x)dx=\frac12$$

Answer 1

As metamorphy points out, we have $?(1-x) = 1 - ?(x)$, so:

$$\int_{0}^{1} ?(x) \text{d}x = \int_{0}^{1} 1-?(1-x) \text{d}x = 1 - \int_{0}^{1}?(1-x)\text{d}x = 1 + \int_{1}^{0}?(u)\text{d}u$$

where $u = 1-x$. If $I = \int_{0}^{1} ?(x)\text{d}x$, then $I = 1 - I$, and we have $I = \frac{1}{2}$.