So I was looking through the homepage of Youtube to see if there were any math problems that I thought that I might be able to solve when I came across this video by Maths$505$ asking how to evaluate this monster integral from the $2023$ MIT Integration Finals. This is the monster floor integral:$$\int_0^1\left(\sum_{n=1}^\infty\frac{\lfloor2^nx\rfloor}{3^n}\right)^2dx$$which I thought that I might be able to evaluate. Here is my attempt at evaluating the monster floor integral:
$$\text{Here's what I know}$$
- From this question of mine, I know that $\lfloor x\rfloor$ can be represented as $x-1$, and that a sum $\left(\sum_{a=\operatorname{Var}[b]}^cd\right)^2$ is equivalent to $\sum_{a=\operatorname{Var}[b]}^cd^3$ (as long as $\operatorname{Var}[b]=1$), we can rewrite the integral as$$\int_0^1\left(\sum_{n=0}^\infty\dfrac{(2^nx-1)^3}{27^n}\right)dx$$Which I realized that I was doing this wrong. I rewrote the integral as$$2\int_0^\frac12\left(\sum_{u\geq2}\dfrac{\lfloor2^{n+1}u\rfloor}{3^u3^{-1}}\right)^2du$$(substituting $x=2u$ and $dx=2du$)$$=18\int_0^\frac12\left(\sum_{u\geq2}\dfrac{\lfloor2^uu\rfloor}{3^u}+\dfrac{\lfloor2u\rfloor}{3}\right)^2du$$$$=18\int_0^\frac12\left(\sum_{u\geq1}\dfrac{2^uu}{3^n}\right)^2du$$Letting $u=t+\dfrac12$ $$I=\int_0^\frac12+\int_\frac12^1\left(\sum_u\dfrac{\lfloor2^ut+2^{u-1}\rfloor}{3^n}\right)^2dt$$$$=\int_0^\frac12+\int_\frac12^1\left(\sum_u\dfrac{\lfloor2^ut\rfloor+2^{u-1}}{3^u}\right)^2dt$$$$=\int_0^\frac12+\int_\frac12^1\left(\sum_u\dfrac{\lfloor2^ut\rfloor}{3^u}+\sum_u\dfrac{2^u}{3^u}\right)^2dt$$Which can be rewritten as$$\int_0^\frac12+\int_\frac12^1\left(\left(\sum_u\dfrac{\lfloor2^ut\rfloor}{3^u}\right)^2+1+2\sum_u\dfrac{\lfloor2^ut\rfloor}{3^u}\right)dt$$$$=2\int_0^\frac12+\int_0^\frac121dt+2\int_0^\frac12\sum_u\dfrac{\lfloor2^ut\rfloor}{3^u}dt$$$$=\sum_{u\geq1}\int_0^\frac12\dfrac{\lfloor2^ut\rfloor}{3^u}dt$$$$=\sum_{u\geq1}\dfrac{1}{3^u}\int_0^\frac12\lfloor2^ut\rfloor dt$$ Letting $$2^ut=x$$$$\implies2^udt=dx$$$$\implies dt=2^{-u}dx$$$$\implies I=\sum_{u\geq1}\dfrac1{3^u}\dfrac1{2^u}\int_0^{2^{u-1}}\lfloor x\rfloor dx$$$$=\sum_{u\geq1}\dfrac1{3^u2^u}\sum_{k=1}^{2^{u-1}-1}k$$$$=\sum_{u\geq1}\dfrac1{3^u2^u}\dfrac{\left(2^{u-1}-1\right)\left(2^{u-1}\right)}2$$$$=\dfrac12\sum_{u\geq1}\dfrac{2^{2u-2}-2^{u-1}}{3^u2^u}$$$$=\dfrac18\sum_{u\geq1}\dfrac{2^u}{3^u}-\dfrac14\sum_{u\geq1}\dfrac{2^{u-1}}{3^u}$$$$=\dfrac14-\dfrac12$$$$=-\dfrac14$$
$$\text{My question}$$
Is my evaluation correct, or what could I do to evaluate it/evaluate it more quickly?
$$\text{Mistakes I might have made}$$
- Evaluating it incorrectly
- Representing it as sums incorrectly
- Incorrect tags on the question
