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Question-193268

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Question Number 193268 by 073 last updated on 09/Jun/23 Answered by qaz last updated on 09/Jun/23 $$\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−{x}\right)^{{n}} }{{n}!}\right)\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−{x}\right)^{{n}} {x}}{{n}!\left({n}+\mathrm{1}\right)}\right)={xe}^{−{x}} \underset{{n}=\mathrm{0}} {\overset{\infty}…

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IS-THIS-RIGHT-I-0-e-ix-2-dx-I-2-0-e-ix-2-dx-0-e-iy-2-dy-0-0-e-iy-2-dye-ix-2-dx-0-0-e-i-x-2-y-2-dydx-dydx-dA-rdrd-I-2-0-pi-2-0-

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Question Number 193316 by gatocomcirrose last updated on 10/Jun/23 $$ \\ $$$$\mathrm{IS}\:\mathrm{THIS}\:\mathrm{RIGHT}? \\ $$$$\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{ix}^{\mathrm{2}} } \mathrm{dx} \\ $$$$\mathrm{I}^{\mathrm{2}} =\int_{\mathrm{0}} ^{\infty} \mathrm{e}^{−\mathrm{ix}^{\mathrm{2}} } \mathrm{dx}\int_{\mathrm{0}}…

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