TITLE: How to prove or disprove $\int_1^\infty \operatorname{arccot}(\cot(\pi x))/(\pi x^2) dx = 1 - \gamma$?
QUESTION [5 upvotes]: Let $\gamma$ be Euler constant.
According to Maple 13:
$$ \int_1^\infty \operatorname{arccot}(\cot(\pi x))/(\pi x^2) dx = 1 - \gamma $$
How to prove or disprove this?
Numerically get approximation which is not
good at high precision, possibly due to numerical
instability.

REPLY [6 votes]: The numerator represents a sawtooth curve, so, the integral can be written as:
\begin{align*}
  \int_{1}^{\infty} \, \frac{\operatorname{arccot}{\cot(\pi x)}}{\pi x^2} \, dx &= \sum_{i=1}^{\infty} \int_{i}^{i+1} \, \frac{x-i}{x^2}\, dx \\
    &= \lim_{n\to\infty} \sum_{i=1}^{n} \log{(i+1)}-\log{i}-\frac{1}{i+1} \\
    &= \lim_{n\to\infty} \log{(n+1)}-H_{n+1} + 1 \\
    &= 1-\gamma
\end{align*}
Update:
In general, for $b>1$ and $b \ne 2$,
$$
\int_{1}^{\infty} \, \frac{\operatorname{arccot}{\cot(\pi x)}}{\pi x^b} \, dx = 
\frac{1}{b-2}-\frac{\zeta(b-1)}{b-1}
$$