Evaluate $\int_0^1 \{\ln{\left(\frac{1}{x}\right)}\} \mathop{dx}$ $$\int_0^1 \left\{\ln{\left(\frac{1}{x}\right)}\right\} \mathop{dx}$$
Where $\{x\}$ is the fractional part of x.  I was wondering if this integral converges and has a closed form but I dont know how to calculate it. I tried $u=\frac{1}{x}$ to get $$\int_1^{\infty} \frac{\{\ln{u}\}}{u^2} \; du$$
and then perhaps convert the numerator into a series somehow...?
 A: Using the change of variable $y = \log(1/x)$, i.e. $x = e^{-y}$, your integral becames
$$
I = \int_0^\infty e^{-y} \{y\}\, dy
= \sum_{n=0}^\infty \int_n^{n+1} e^{-y}(y-n)\, dy
= \sum_{n=0}^\infty e^{-n} (1 - 2/e) = \frac{e-2}{e-1}.
$$
A: Just to give a slightly different take, let $x=e^{-(n+r)}$ with $n\in\mathbb{N}$ and $0\le r\lt1$. Then
$$\int_0^1\{\ln(1/x)\}dx=\sum_{n=0}^\infty e^{-n}\int_0^1re^{-r}dr={e\over e-1}\int_0^1re^{-r}dr$$
and
$$\begin{align}
\int_0^1re^{-r}dr
&=\int_0^\infty re^{-r}dr-\int_1^\infty r^{-r}dr\\
&=\Gamma(2)-\int_0^\infty(u+1)e^{-(u+1)}du\\
&=\Gamma(2)-{1\over e}(\Gamma(2)+\Gamma(1))\\
&=1-{1\over e}(1+1)\\
&={e-2\over e}
\end{align}$$
so
$$\int_0^1\{\ln(1/x)\}dx={e-2\over e-1}$$
A: Make the change of variable $x=\exp{(-t)}$ to obtain
\begin{align} I:=\int_0^1\left\{\ln{\left(\frac{1}{x}\right)}\right\}\:dx&=\int_0 ^{\infty}\{t\}\cdot \exp{(-t)}\:dt \\ &= \int_0 ^{\infty}(t-\lfloor{t}\rfloor)\cdot \exp{(-t)}\:dt \\ &=\int_0^{\infty}t\cdot\exp{(-t)}\:dt-\sum_{k=0}^{\infty}\ \left\{\int_k^{k+1}k\cdot \exp(-t)\:dt\right\} \\ &= 1-\sum_{k=0}^{\infty}\ k\cdot(\exp(-k)-\exp(-k-1)) \\ &=1-\frac{1}{e-1} \\ &= \frac{e-2}{e-1}\end{align}
The problem is solved.
