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Question Number 129376 by Dwaipayan Shikari last updated on 15/Jan/21

$${\int }_0^1{e}^{-x^2}dx=\frac{1}{1+\frac{1}{2+\frac{3^2}{7+\frac{{10}^2}{32+\frac{{42}^2}{174+\frac{{216}^2}{1196+\frac{{1312}^2}{....}}}}}}}$$

Commented by Dwaipayan Shikari last updated on 15/Jan/21

$${\int }_0^a{e}^{-x^2}dx={\int }_0^{a}1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+\frac{x^8}{4!}-...$$$$=a-\frac{a^3}{3.1!}+\frac{a^5}{5.2!}-\frac{a^7}{7.3!}+\frac{a^9}{9.4!}-..$$$$=a+a(-\frac{a^2}{3.1!})+a.\left(\frac{-a^2}{3.1!}\right)\left(\frac{-3.1!a^2}{5.2!}\right)+a\left(\frac{-a^2}{3.1!}\right)(-\frac{3.1!a^2}{5.2!})\left(\frac{-5.2!}{7.3!}{a}^2\right)+..$$$$=\frac{a}{1+\frac{\frac{a^2}{3.1!}}{1-\frac{a^2}{3.1!}+\frac{\frac{3.1!a^2}{5.2!}}{1-\frac{3.1!a^2}{5.2!}+\frac{\frac{5.2!a^2}{7.3!}}{1-\frac{5.2!}{7.3!}{a}^2+...}}}}$$$$=\frac{a}{1+\frac{a^2}{3.1!-a^2+\frac{{(3.1!)}^2{a}^2}{5.2!-3.1!a^2+\frac{{(5.2!)}^2{a}^2}{7.3!-5.2!a^2+{(7.3!)}^2...}}}}$$$$$$

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