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

Question

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.

Answer 1

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 $$ I_N=\log(2\pi)-(2N+1)\log(1+1/N)+o(1)=\log(2\pi)-2+o(1). $$

Answer 2

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.