İntegrali parçalara ayıralım. $$I=\int_{1}^{\infty} \dfrac{\{(-1)^{\lfloor x\rfloor}\cdot x\}}{x^2} dx=\sum_{k=1}^{\infty} \int_{k}^{k+1} \dfrac{\{(-1)^{\lfloor x\rfloor}\cdot x\}}{x^2} dx=\sum_{i=1}^{\infty} \int_{2i}^{2i+1} \dfrac{\{(-1)^{\lfloor x\rfloor}\cdot x\}}{x^2} dx+\sum_{i=1}^{\infty} \int_{2i-1}^{2i} \dfrac{\{(-1)^{\lfloor x\rfloor}\cdot x\}}{x^2} dx$$ $$\Rightarrow I=\sum_{i=1}^{\infty} \int_{2i}^{2i+1} \dfrac{\{ x\}}{x^2} dx+\sum_{i=1}^{\infty} \int_{2i-1}^{2i} \dfrac{\{-x\}}{x^2} dx \\ =\sum_{i=1}^{\infty} \int_{2i}^{2i+1} \dfrac{\{ x\}}{x^2} dx+\sum_{i=1}^{\infty} \int_{2i-1}^{2i} \dfrac{1-\{x\}}{x^2} dx=\sum_{i=1}^{\infty} \int_{2i}^{2i+1} \dfrac{x-2i}{x^2} dx+\sum_{i=1}^{\infty} \int_{2i-1}^{2i} \dfrac{1-x+(2i-1)}{x^2} dx$$ Bu integrali alıp toplam sembollerini birleştirirsek, $$\Rightarrow I=\sum_{i=1}^{\infty}\left (\ln \left (\dfrac{(2i+1)(2i-1)}{(2i)^2}\right )+\left (\dfrac{1}{2i-1}-\dfrac{1}{2i+1}\right ) \right )=\ln \left (\prod_{i=1}^{\infty} \left (\dfrac{(2i+1)(2i-1)}{(2i)^2}\right ) \right )+1$$ $\dfrac{2\cdot 2\cdot 4\cdot 4\cdot 6\cdot 6 \cdots}{1\cdot 3\cdot 3\cdot 5\cdot 5\cdot 7 \cdots}=\dfrac{\pi}{2}$ eşitliğini yazarsak, $$I=\ln \left (\dfrac{2}{\pi}\right)+1$$ bulunur.
