Integral of the fractional part of $\frac1x$ multiplied by $x$ on interval $(a,b), a\ge 0$.

Question

I'm interested in finding the value of

1. the integral of $\left\{\frac{1}{x}\right\}\cdot x$ (the fractional part of $\dfrac{1}{x}$ multiplied by $x$) on the interval $(a,b), a\ge 0$
2. the integral of $\left\{\frac{1}{x}\right\}$ (fractional part of $\dfrac{1}{x}$) on the interval $(a,b), a\ge 0$

NOTE: $\left\{x \right\}= x-\left\lfloor x \right\rfloor $

Thanks

Answer 1

A hint: Your integrands misbehave at the points $x_k$ where $1/x$ is an integer $k$. Therefore split each of the integrals (1) and (2) up in a sum of integrals over intervals $[x_{k+1},x_k]$.

Answer 2

For positive integers $n$, $$\int_{1/(n+1)}^{1/n} x \text{frac}(1/x)\ dx = \int_{1/(n+1)}^{1/n} x (1/x - n)\ dx = \frac{1}{2n(n+1)^2}$$ so $$\int_0^{1/n} x \text{frac}(1/x)\ dx = \sum_{j=n}^\infty \frac{1}{2j(j+1)^2} = \frac{1}{2n} - \frac{\Psi(1,n+1)}{2}$$ If $b > 0$ and $n = \lfloor 1/b \rfloor$, $$\int_0^b x \text{frac}(1/x)\ dx = \frac{1}{2(n+1)} - \frac{\Psi(1,n+2)}{2} + \int_{1/(n+1)}^b x (1/x - n-1)\ dx = \frac{1 - (b(n+1)-1)^2}{2(n+1)} - \frac{\Psi(1,n+2)}{2}$$ and $\int_a^b x \text{frac}(1/x)\ dx = \int_0^b x \text{frac}(1/x)\ dx - \int_0^a x \text{frac}(1/x)\ dx$.