Javob: $\sqrt{\pi}$. Yechim: $I = \int_{-\infty}^\infty e^{-x^2} dx$ ni quyidagicha hisoblashimiz mumkin: 1. Avval integralni kvadratga ko'taramiz: $I^2 = \left{-\int_{-\infty}^\infty e^{-x^2} dx \right) ^2 = \int_{-\infty}^\infty \int_{-\infty}^\infty {-x^2} dx dy.$ 2. Polar koordinatalarga o'tish. Eslab qolingki, $d dyx = d d \theta dr$: $I2 = \int_^0^\infty \int_^2{-\infty} e^{-
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