How to evaluate $\int_0 ^1 \left( \sum\limits_{n=1}^{\infty} \frac{\lfloor2^nx \rfloor}{3^n}\right)^2dx$?

Question

I want to evaluate $$\displaystyle \int_0 ^1 \left( \sum\limits_{n=1}^{\infty} \frac{\lfloor2^nx \rfloor}{3^n}\right)^2dx$$

I have no idea how to solve this integral

I have tried expressing $x $ as $\sum\limits_{n=1}^\infty (-1)^{f(n)}\frac{1}{2^n}$ such that $\{f(x)\}=\{0,1\} $ but it became a very messy formula that I think it is impossible to evaluate.

Since $$ \frac{1}{3^{2m-2}}\le \int_{2^{-m}} ^{2^{-m+1}} \left( \sum\limits_{n=1}^{\infty} \frac{\lfloor2^nx \rfloor}{3^n}\right)^2dx \le \frac{1}{3^{2m-4}} $$ I have tried using $u = 2^m x$ for some "big" $m$ but the problem is the interval of integration will be $[0, 2^m]$

Answer 1

I am reproducing my quora answer verbatim.

I am not sure how someone can do this under 4 minutes! Initially I tried expanding the square into a double sum & pulling the integral inside, but it turned out to be extremely tedious. Here's an approach that works: $\displaystyle \begin{align*} \mathcal{I} & =\ \int _{0}^{1}\left(\sum _{n=1}^{\infty }\frac{\lfloor 2^{n} x\rfloor }{3^{n}}\right)^{2}\mathrm{d} x\\ & =2\int _{0}^{\frac{1}{2}}\left(\sum _{n=1}^{\infty }\frac{\lfloor 2^{n+1} x\rfloor }{3^{n}}\right)^{2}\mathrm{d} x\tag{01}\\ & =2\int _{0}^{\frac{1}{2}}\left( 3\cdotp \sum _{n=1}^{\infty }\frac{\lfloor 2^{n+1} x\rfloor }{3^{n+1}}\right)^{2}\mathrm{d} x\\ & =18\underbrace{\int _{0}^{\frac{1}{2}}\left(\sum\limits _{n=1}^{\infty }\frac{\lfloor 2^{n} x\rfloor }{3^{n}}\right)^{2}\mathrm{d} x}_{\mathcal{J}}\tag{02} \end{align*}$ Also, we can split $\mathcal I$ at the point $x=1/2$. $\displaystyle \begin{align*} \mathcal{I} & =\mathcal{J} +\int _{\frac 12}^{1}\left(\sum\limits _{n=1}^{\infty }\frac{\lfloor 2^{n} x\rfloor }{3^{n}}\right)^{2}\mathrm{d} x\\ & =\mathcal{J} +\int _{0}^{\frac 12}\left(\sum\limits _{n=1}^{\infty }\frac{\lfloor 2^{n} x+2^{n-1} \rfloor }{3^{n}}\right)^{2}\mathrm dx\tag{03}\\ & =\mathcal{J} +\int _{0}^{\frac 12}\left(\sum\limits _{n=1}^{\infty }\frac{\lfloor 2^{n} x\rfloor }{3^{n}} +\frac 13\cdot\sum\limits _{n=0}^{\infty }\left(\frac{2}{3}\right)^{n}\right)^{2}\mathrm dx\\ & =\mathcal{J} +\int _{0}^{\frac 12}\left(\sum\limits _{n=1}^{\infty }\frac{\lfloor 2^{n} x\rfloor }{3^{n}} +1\right)^{2}\mathrm dx\\ & =2\mathcal{J} +2\underbrace{\int _{0}^{\frac 12}\sum\limits _{n=1}^{\infty }\frac{\lfloor 2^{n} x\rfloor }{3^{n}}\mathrm{d} x}_{\mathcal{K}} +\frac{1}{2}\\ & =\frac{\mathcal{I}}{9} +2\mathcal{K} +\frac{1}{2} \tag{04}\\ &=\frac 98\left(2\mathcal K +\frac 12\right) \end{align*}$ Now $\mathcal K$ is relatively easy to solve. $\displaystyle \begin{align*}\mathcal K &= \int _{0}^{\frac 12}\sum\limits _{n=1}^{\infty }\frac{\lfloor 2^{n} x\rfloor }{3^{n}}\mathrm{d} x\\ &=\sum_{n=1}^\infty\frac{1}{3^n}\int_{0}^{\frac 12}\lfloor 2^n x\rfloor \mathrm dx\\ &= \sum_{n=1}^\infty\frac{1}{3^n\cdot 2^n}\int_0^{2^{n-1}}\lfloor x\rfloor\mathrm dx\tag{05}\\ &= \sum_{n=1}^\infty\frac{1}{2^n\cdot 3^n}\cdot \frac{2^{n-1}\left(2^{n-1}-1\right)}{2}\tag{06}\\ &=\frac 1{12}\sum_{n=1}^\infty\left\{\left(\frac 23\right)^{n-1}-\left(\frac{1}{3}\right)^{n-1}\right\}\\ &=\frac{1}8\end{align*}$ $\therefore \displaystyle \mathcal I = \frac 98\left(2\mathcal K +\frac 12\right)=\boxed{\frac {27}{32}}\tag*{}$

Explanation of the tagged steps:

1. Replace $x$ by $2x$.
2. This step is justified because $\lfloor 2^1x \rfloor=0 \ \forall \ x\in[0,1/2)$
3. Replace $x$ by $x+1/2$.
4. $\because \mathcal I = 18\mathcal J$ from $(02)$.
5. Replace $x$ by $2^{-n}x$.
6. $\displaystyle \because \int_0^n \lfloor x \rfloor \mathrm dx = \frac{n(n-1)}{2} \ \forall \ n \in \mathbb N$

Also note that there are several instances where I've applied the sum of infinite geometric series. $\displaystyle \forall \ |r|<1, \ \sum_{n=0}^\infty r^n=\frac{1}{1-r}\tag*{}$