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How do I integrate $$\int_{0}^1 x \bigg\lceil \frac{1}{x} \bigg\rceil \left\{ \frac{1}{x} \right\}\, dx$$

Where $\lceil x \rceil $ is the ceiling function, and $\left\{x\right\}$ is the fractional part function

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A related problem.

**Hint:** try to use the definition of the fractional part function which is defined by

$$ \left\{ x\right\} = x – \lfloor x\rfloor , $$

and the following relation between the floor and ceiling functions

$$ \lceil x \rceil – \lfloor x \rfloor = \begin{cases} 0&\mbox{ if } x\in \mathbb{Z}\\ 1&\mbox{ if } x\not\in \mathbb{Z} \end{cases}. $$

**Added:**

$$ \int_{0}^1 x \bigg\lceil \frac{1}{x} \bigg\rceil \left\{ \frac{1}{x} \right\}\, dx=\int_{0}^1 x (1+\lfloor 1/x \rfloor)(1/x-\lfloor1/x\rfloor)\, dx. $$

Now, make the change of variables $y=1/x$ to the last integral

$$\int_{0}^1 x (1+\lfloor 1/x \rfloor)(1/x-\lfloor1/x\rfloor)\, dx=\int_{1}^{\infty} \frac{1}{y} (1+\lfloor y \rfloor)(y-\lfloor y\rfloor)\, \frac{dy}{y^2}$$

$$\implies I = \sum_{n=1}^{\infty}\int_{n}^{n+1} \frac{1}{y^3} (1+n)(y-n)\, dy= \frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n(n+1)}=\frac{1}{2} $$

**Note:** To evaluate the sum, use the telescoping technique. First write the summand as

$$ \frac{1}{n(n+1)}= \frac{1}{n}-\frac{1}{n+1}. $$

Now, find the partial sum of the series

$$ s_n = \sum_{k=1}^{n} \left( \frac{1}{k}-\frac{1}{k+1} \right)=1-\frac{1}{n+1}. $$

Then the series sums to

$$ s = \lim_{n \to \infty} s_n = 1. $$

The main idea is to divide $(0,1)$ into “good” intervals. I’lll give only the main steps of computation

$$

\int\limits_{(0,1)} x \bigg\lceil \frac{1}{x} \bigg\rceil \left\{ \frac{1}{x} \right\} dx

=\sum\limits_{n=1}^\infty\int\limits_{n\leq \frac{1}{x}<n+1} x \bigg\lceil \frac{1}{x} \bigg\rceil \left\{ \frac{1}{x} \right\} dx

=\sum\limits_{n=1}^\infty\int\limits_{\frac{1}{n+1}< x\leq \frac{1}{n}} x (n+1) \left(\frac{1}{x}-n\right) dx

=\sum\limits_{n=1}^\infty\frac{1}{2n^2+2n}=\frac{1}{2}\sum\limits_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{n+1}\right)=\frac{1}{2}

$$

Split the integral up into segments $S_m=[1/m,1/(m+1)]$ with $[0,1]= \cup_{m=1}^\infty S_m$. In the segment $m$, we have $\lceil 1/x \rceil=m+1$ and $\{1/x\} = 1/x- \lfloor 1/x\rfloor = 1/x – m$ (apart from values of $x$ on the boundary which do not contribute to the integral).

This yields

$$\begin{align}\int_0^1 x \bigg\lceil \frac{1}{x} \bigg\rceil \left\{ \frac{1}{x} \right\}\, dx &= \sum_{m=1}^\infty \int_{S_m}x \bigg\lceil \frac{1}{x} \bigg\rceil \left\{ \frac{1}{x} \right\}\, dx \\

&= \sum_{m=1}^\infty \int_{1/(m+1)}^{1/m} x (m+1)\left(\frac1x -m \right)\, dx\\

&= \sum_{m=1}^\infty \frac{1}{2m(1+m)}\\

&=\frac{1}{2}.

\end{align}$$

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