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.

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.  
