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: enter image description here

This result seems correct for a closed form with the exponential integral function:enter image description here

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:

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

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    $\begingroup$ $?(1-x)=1-{?}(x)$ immediately gives $\int_0^1 ?(x)\,dx=1/2$. $\endgroup$
    – metamorphy
    Jul 6, 2021 at 3:24
  • $\begingroup$ ?(x) is a singular continuous function. Its derivative is 0 a.e. Approximations to the derivative are interesting. $\endgroup$ Sep 24, 2021 at 0:37
  • $\begingroup$ @YuvalPeres Please see this other integral of “fractal” function. Do you know if it is possible to find another definite or indefinite integral of $ ?(x)$ $\quad$? $\endgroup$ Sep 24, 2021 at 0:41

1 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 $I = \frac{1}{2}$.

  • $\begingroup$ Good job!. What would happen if we let 1=x. We can use this method to get the integral of ?(x). $\endgroup$ Jul 6, 2021 at 11:37
  • $\begingroup$ I'm not understanding exactly what you mean. If $x=1$, this gives you the value of $?(x)$. Are you looking to get a general antiderivative of $?(x)$ ? $\endgroup$
    – Moni145
    Jul 6, 2021 at 15:55
  • $\begingroup$ Yes, I have tried it out, and it seems that the same technique cannot be used as the integrals cannot be combined after substitution. I wonder if we can find a general anti derivative of “?(x)”. I saw this post too. $\endgroup$ Jul 6, 2021 at 16:01

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