For $0<x<1$, let $f(x)=\int_0^1\left( \left\lfloor\frac{x}{y}\right\rfloor-x \left\lfloor\frac{1}{y}\right\rfloor\right)dy$, $\lfloor. \rfloor$ denotes the greatest integer function. If $f(x)$ attains local maximum at $x=k$, where $(0<k<1)$ then value of $\dfrac{3ek}{2}$
My Attempt: All i thought is I replaced floor with ceil funtion. After that I cannnot proceed further at all.
Well, I have doubt. Note that we can't use the linearity of integral to open the brackets in the original question because we have $$ \int^{1}_{0}\left\lfloor\dfrac{x}{y}\right\rfloor{\rm d}y = \sum^{\infty}_{k=1}\int^{\frac{x}{k}}_{\frac{x}{k+1}}\left\lfloor\dfrac{x}{y}\right\rfloor{\rm d}y = \sum^{\infty}_{k=1}k\left(\dfrac{x}{k}-\dfrac{x}{k+1}\right) = \sum^{\infty}_{k=1}\dfrac{x}{k+1}=\infty, $$ so opening the brackets would result in evaluating $\infty-\infty$. However, we can consider the fractional parts $\{x\}=x-\lfloor x\rfloor$ and write $$ f(x) = \int^{1}_{0}\left(x\left\{\dfrac{1}{y}\right\}-\left\{\dfrac{x}{y}\right\}\right){\rm d}y, $$ then $f(x)$ would be the difference of two well-defined integrals. In fact, if we pose $u(x) = \displaystyle\int^{1}_{0}\left\{\dfrac{x}{y}\right\}{\rm d}y$, then we have \begin{align*} u(x) &= \int^{1}_{x}\left\{\dfrac{x}{y}\right\}{\rm d}y+\sum^{\infty}_{k=1}\int^{\frac{x}{k}}_{\frac{x}{k+1}}\left\{\dfrac{x}{y}\right\}{\rm d}y = \int^{1}_{x}\dfrac{x}{y}{\rm d}y+\sum^{\infty}_{k=1}\int^{\frac{x}{k}}_{\frac{x}{k+1}}\left(\dfrac{x}{y}-k\right){\rm d}y\\&=-x\log x+Cx,\quad\forall x\in[0,1], \end{align*} where $0\log 0$ is understood as $0$, and $C:=\displaystyle\sum^{\infty}_{k=1}\left(\log\left(1+\dfrac{1}{k}\right)-\dfrac{1}{k+1}\right)$ is a well-defined constant because $\log\left(1+\dfrac{1}{k}\right)-\dfrac{1}{k+1}\sim\dfrac{1}{2k^2}$ as $k\to\infty$. (We have $C=1-\gamma$, where $\gamma$ is the Euler-Mascheroni constant; see OEIS A153810.) So we have $$ f(x) = xu(1) - u(x) = x\log x,\quad\forall x\in[0,1], $$ which attains its minimum value $-\dfrac{1}{e}$ at $x=\dfrac{1}{e}$.