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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.

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  • $\begingroup$ does this mean Maple is better than Mathematica? $\endgroup$ – Santosh Linkha Jun 3 '14 at 16:23
  • $\begingroup$ @SantoshLinkha don't know, what Mathematica says about it? $\endgroup$ – elluser Jun 3 '14 at 16:36
  • $\begingroup$ Mathematica can't compute it ... probably due to $\cot(\pi x)$ term there. $\endgroup$ – Santosh Linkha Jun 3 '14 at 16:38
  • $\begingroup$ @SantoshLinkha I suspect this is true, but don't see how Maple derived this... $\endgroup$ – elluser Jun 3 '14 at 16:40
  • $\begingroup$ Mathematica 9.0.1.0 states the integral under consideration diverges. This is not true because the numerator of the integrand is nonnegative and less than or equal to $\pi$. The Maple code $$ int(arccot(cot(Pix))/(Pix^2), x = 1 .. 1000, numeric, epsilon = 0.1e-2)$$ outputs $0.4222844184 $, confirming $1-\gamma .$ $\endgroup$ – user64494 Jun 3 '14 at 16:47
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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 $b \ne 2$,

$$ \int_{1}^{\infty} \, \frac{\operatorname{arccot}{\cot(\pi x)}}{\pi x^b} \, dx = \frac{1}{b-2}-\frac{\zeta(b-1)}{b-1} $$

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  • $\begingroup$ Thank you. Does this work for denominators like $\pi x^4$ or $\pi x^{3/2}$? $\endgroup$ – elluser Jun 3 '14 at 17:25
  • $\begingroup$ Yes, it does. When denominator is $\pi x^4$, integral is $\displaystyle \frac{1}{2}-\frac{\zeta(3)}{3}$, I was trying for the closed form for the $\pi x^{3/2}$, couldn't get it yet. $\endgroup$ – gar Jun 3 '14 at 17:54
  • $\begingroup$ For s>-1 the integral with denominator \pi x^{s+1} is simple function of \zeta(s). $\endgroup$ – elluser Jun 4 '14 at 5:13
  • $\begingroup$ For s>2, it's easy to see that $$\int_{1}^{\infty} \, \frac{\operatorname{arccot}{\cot(\pi x)}}{\pi x^2} \, dx = \frac{1}{s-2}-\frac{\zeta(s-1)}{s-1}$$ For $2>s>1$, is a bit difficult for me. $\endgroup$ – gar Jun 4 '14 at 6:02
  • $\begingroup$ gar, this is related to the fractional part {x} of x. Check the identity and use the integral for zeta: math.stackexchange.com/questions/820111/… $\endgroup$ – elluser Jun 4 '14 at 6:04

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