\frac{3}{4 \pi} \sqrt{4 \cdot x^2 12}\\ \lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k^2} = \frac{\pi^2}{6}\\ \it{f}(x) = \frac{1}{\sqrt{x} x^2}\\ e^{i \pi} + 1 = 0}\; m^2~ha^{-1}