Simple integral help 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

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

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

