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I am interested in a proof of the following.

$ \int_1^{\infty} \dfrac{\{t\} (\{t\} - 1)}{t^2} dt = \log \left(\dfrac{2 \pi}{e^2}\right)$ where $\{t\}$ is the fractional part of $t$.

I obtained a circuitous proof for the above integral. I'm curious about other ways to prove the above identity. So I thought I will post here and look at others suggestion and answers.

I am particularly interested in different ways to go about proving the above.

I'll hold off from posting my proof for sometime to see what all different proofs I get for this.

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    Any time I see an integral involving a function that's piecewise well-behaved, my first instinct is to break it up into well-behaved pieces. Unfortunately, my series-fu is weak, and I don't know how to sum the series after that. Mathematica gets the right sum but it doesn't have a "show steps" button...2012-06-08

2 Answers 2

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The integral on $[1,N+1]$ is (see @Rahul's first comment) $ I_N=\sum_{n=1}^N\big(2+2\log n+(2n-1)\log n-(2n+1)\log(n+1)\big), $ that is, $ I_N=2N+2\log(N!)-(2N+1)\log(N+1). $ Thanks to Stirling's approximation, $2\log(N!)=(2N+1)\log N-2N+\log(2\pi)+o(1)$. After some simplifications, this leads to $ I_N=\log(2\pi)-(2N+1)\log(1+1/N)+o(1)=\log(2\pi)-2+o(1). $

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    @Marvis: http://math.stackexchange.com/questions/159018/evaluating-lim-n-to-infty-frac-sqrt-n-sqrt-2n-int-0-frac-pi/159029#comment366657_1590292012-06-16
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Let's consider the following way involving some known results of celebre integrals with fractional parts:

$ \int_1^{\infty} \dfrac{\{t\} (\{t\} - 1)}{t^2} dt = \int_1^{\infty} \dfrac{\{t\}^2}{t^2} dt - \int_1^{\infty} \dfrac{\{t\}}{t^2} dt = \int_0^1 \left\{\frac{1}{t}\right\}^2 dt- \int_0^1 \left\{\frac{1}{t}\right\} dt = (\ln(2\pi) -\gamma-1)-(1-\gamma)=\ln(2\pi)-2=\log \left(\dfrac{2 \pi}{e^2}\right).$

REMARK: there is a theorem that establishes a way of calculating the value of the below integral for $m\geq1$:

$\int_0^1 \left\{\frac{1}{x}\right\}^m dx$ enter image description here

The proof is complete.

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    @did: you're welcome any time.2012-06-08