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$ \int_{-\infty}^{\infty} e^{-x^{2}} d x=\sqrt{\pi} $

$ 1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\ldots \ldots=\frac{\pi^{2}}{6} $

$ e^{i \pi}+1=0 $

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