Consider $\int_0^1 (\lfloor\frac{x}{y}\rfloor-x\lfloor\frac{1}{y}\rfloor) dy$, where $0\lt x\lt 1$ ...

Question

Consider $f(x)=\int_0^1 (\lfloor\frac{x}{y}\rfloor-x\lfloor\frac{1}{y}\rfloor) dy$ where $0<x<1$

$\implies \int_0^1 \lfloor\frac{x}{y}\rfloor dy - \int_0^1x\lfloor\frac 1y\rfloor dy$

Let the integrals be $I_1$ and $I_2$ respectively

$\therefore, I_1 = \int_x^1 \lfloor\frac{x}{y}\rfloor dy + \int_{x/2}^{x} \lfloor\frac{x}{y}\rfloor dy + \int_{x/3}^{x/2} \lfloor\frac{x}{y}\rfloor dy \cdots $

$\implies I_1= \int_x^1 0.dy + \int_{x/2}^{x} 1.dy + \int_{x/3}^{x/2} 2.dy \cdots $

$\implies I_1= 0+\frac{x}{2}+\frac{x}{3}+\frac{x}{4} \cdots$

Similarly, $I_2 = \int_{1/2}^1x\lfloor\frac{1}{y}\rfloor dy + \int_{1/3}^{1/2}x\lfloor\frac{1}{y}\rfloor dy+\int_{1/4}^{1/3}x\lfloor\frac{1}{y}\rfloor dy \cdots$

$\implies I_2= \frac{x}{2}+\frac{x}{3}+\frac{x}{4} \cdots $

$\therefore, I_1 - I_2=0$ for all $x$

However this question on math.stack has a different answer.

So, where is my solution incorrect?

Answer 1

I have already linked the answer to the question in the comments.

Here, I would tell you what mistake you did while solving. $\therefore, I_1 = \int_x^1 \lfloor\frac{x}{y}\rfloor dy + \int_{x/2}^{x} \lfloor\frac{x}{y}\rfloor dy + \int_{x/3}^{x/2} \lfloor\frac{x}{y}\rfloor dy \cdots \implies I_1= 0 + 1+2+ \cdots$

which is not true. Instead it should be $$\therefore, I_1 = \int_x^1 \lfloor\frac{x}{y}\rfloor dy + \int_{x/2}^{x} \lfloor\frac{x}{y}\rfloor dy + \int_{x/3}^{x/2} \lfloor\frac{x}{y}\rfloor dy \cdots \implies \int_x^1 0.dy + \int_{x/2}^{x} 1.dy + \int_{x/3}^{x/2} 2.dy \cdots$$

Same mistake was repeated in $I_2$ too.

Edit :

Now,we can observe that both $I_1$and $I_2$ are both tending to $\infty$. Thus the step,

$\therefore I_1 - I_2=0$ for all $x$.

is not valid as $I_1 - I_2$ is of the form $\infty-\infty$ which is an indeterminate form. You can check about that here:https://en.wikipedia.org/wiki/Indeterminate_form