Convert binary to decimal

We split this integral at $x=\dfrac{1}{2}$ to get: $\displaystyle I=\int\limits_0^{\frac{1}{2}} f(x)dx+\int\limits_0^{\frac{1}{2}} f\left(\dfrac{1}{2}+x\right)dx$

Now, if $0\le x<\dfrac{1}{2}$ then $x$ has the form $x=(0.0a_2a_3\ldots)_2$ , and

$\qquad\;\;\, f(x)=(0.0a_2a_3\ldots)_{10}=\dfrac{1}{10}(0.a_2a_3\ldots)_{10}$

$\qquad\qquad\quad=\dfrac{1}{10}f((0.a_2a_3\ldots)_2)=\dfrac{1}{10}f(2x)$

$f\left(\dfrac{1}{2}+x\right)=(0.1a_2a_3\ldots)_{10}=\dfrac{1}{10}+\dfrac{1}{10}(0.a_2a_3\ldots)_{10}$

$\qquad\qquad\quad=\dfrac{1}{10}+\dfrac{1}{10}f((0.a_2a_3\ldots)_2)=\dfrac{1}{10}+\dfrac{1}{10}f(2x)$

Hence,

$\displaystyle I=\dfrac{1}{10}\int\limits_0^{\frac{1}{2}}\left(f(2x)+1+f(2x)\right)dx=\dfrac{1}{20}\int\limits_0^1\left(1+2f(y)\right)dy=\dfrac{1}{20}+\dfrac{I}{10}$

Thus, $I=\dfrac{1}{18}$ .

So, $m+n=\boxed{19}$ .

Nice problem!

Let's see what $f(x)$ is - Converting to binary involves running it through ${2}^{k}$ s and seeing where it becomes $1$ or $0$ and then dividing each term by its corresponding ${10}^{-k}$ s

So, we can write

$\displaystyle f(x) = \sum_{k=1}^{\infty}{\frac{\left \lfloor {2}^{k} x \right \rfloor mod 2}{{10}^{k}}}$

Now, we need to integrate

$\displaystyle \int_{0}^{1}{\sum_{k=1}^{\infty}{\frac{\left \lfloor {2}^{k} x \right \rfloor mod 2}{{10}^{k}}} dx}$

$\displaystyle \sum_{k=1}^{\infty}{\frac{1}{{10}^{k}} \int_{0}^{1}{\left \lfloor {2}^{k} x \right \rfloor mod 2 \quad dx}}$

Claim : $\displaystyle \quad \int_{0}^{1}{\left \lfloor {a}^{k} x \right \rfloor mod a \quad dx} = \frac{a-1}{2}$ where $a$ is a positive integer.

Proof :

$\displaystyle \frac{1}{{a}^{k}} \int_{0}^{{a}^{k}}{\left \lfloor u \right \rfloor mod a \quad du}$

The integral can be done with the help of graph.

Graph for $\left \lfloor u \right \rfloor$ is simple to draw. An increasing graph with critical points at integers. Now, graph for $\left \lfloor u \right \rfloor mod a$ is a periodic one with period $a$ .

Hence, the area under such a graph for the interval $(0, {a}^{k})$ would be -

$\text{Area for (0,a)}*{a}^{k-1}$ and as in the given interval, there are $a$ rectangles with width $= 1$ and length $= 0, 1, 2, \cdots, a-1$ .

$\displaystyle = {a}^{k-1} \sum_{m=0}^{a-1}{m*1} = \frac{{a}^{k}(a-1)}{2}$

And there is a factor we've not considered, putting it back -

$\displaystyle \frac{a-1}{2}$

Back to the integral, we apply this and get -

$\displaystyle \sum_{k=1}^{\infty}{\frac{1}{{10}^{k}} \frac{1}{2}}$

$\displaystyle \boxed{\frac{1}{18}}$