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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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    Well, for $t$ between integers $n$ and $n+1$, the integral becomes $\int_n^{n+1} (t-n)(t-n-1)/t^2\, dt = 2 + (2n+1)(\log n - \log(n+1))$. So the problem is not really one of integration but of summing that series from $n=1$ to $\infty$, right?2012-06-08
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    @RahulNarain Yes. Right.2012-06-08
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    @RahulNarain Just curious. What made you split the integrals from $n$ to $n+1$ and sum them up? Did you find the pattern by trying to integrate the above with mathematica by taking the upper limit as a large value (or) did the $\{t\}$ influence you to split the integral from $n$ to $n+1$ and sum it up? I am asking this question to understand how people think and go about doing a problem.2012-06-08
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    It's simple: $\{t\}$ is a piecewise linear function of $t$, and the pieces are $[n,n+1)$.2012-06-08
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    @RahulNarain Thanks. The reason why I asked is that I tried in mathematica to do the integral directly but mathematica had issues computing the integral. Hence before posting the question, I tried by taking large values for the upper limit and found what you have in your first comment. Though now thinking about it, it seems natural to split the integral as sum of integrals between two consecutive integers since $\{t\}$ is periodic. :-)2012-06-08
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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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    +1. This was what I did. In some sense, I would like to know what are all the different ways to find the constant $\sqrt{2 \pi}$. Couple of methods, I know are based on the saddle point method and another one is a bit more elaborate computation of the asymptotic.2012-06-09
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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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    Which book is this taken from?2012-06-08
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    @did: http://www.m-hikari.com/ijcms-2011/13-16-2011/luyouminIJCMS13-16-2011.pdf2012-06-08
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    I see. Thanks. $ $2012-06-08
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    @did: you're welcome any time.2012-06-08