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Question Number 64579 by Tawa1 last updated on 19/Jul/19

$$\int {\mathrm{e}}^{{\mathrm{x}}^2}\mathrm{dx}$$$$$$$$\mathrm{can}\mathrm{we}\mathrm{get}\mathrm{a}\mathrm{close}\mathrm{form}\mathrm{of}\mathrm{this}\mathrm{integral}\mathrm{or}\mathrm{analytic}\mathrm{solution}$$

Commented by mathmax by abdo last updated on 19/Jul/19

$$atformofserie$$$$\int e^{x^2}dx=\int \sum _{n=0}^{\mathrm{\infty }}\frac{x^{2n}}{n!}dx=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}\int x^{2n}dx$$$$=\sum _{n=0}^{\mathrm{\infty }}\frac{x^{2n+1}}{(2n+1)n!}+cwithradiusofconvergenceR=+\mathrm{\infty }$$

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