# Evaluating $\int_1^{\infty} \frac{\{t\} (\{t\} - 1)}{t^2} dt$

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? –  Rahul Jun 8 '12 at 3:46
@RahulNarain Yes. Right. –  user17762 Jun 8 '12 at 3:50
@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. –  user17762 Jun 8 '12 at 4:15
It's simple: $\{t\}$ is a piecewise linear function of $t$, and the pieces are $[n,n+1)$. –  Rahul Jun 8 '12 at 4:24
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... –  Rahul Jun 8 '12 at 4:32

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. –  user17762 Jun 9 '12 at 17:59
–  Chris's sis Jun 16 '12 at 11:52
@Chris Thanks. ${}$ –  user17762 Jun 16 '12 at 23:32

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 $m\geq1$:

$$\int_0^1 \left\{\frac{1}{x}\right\}^m dx$$

The proof is complete.

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Which book is this taken from? –  Did Jun 8 '12 at 8:49
–  Chris's sis Jun 8 '12 at 8:56
I see. Thanks.  –  Did Jun 8 '12 at 8:59
@did: you're welcome any time. –  Chris's sis Jun 8 '12 at 9:00