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

$$\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{−{x}^{\mathrm{2}} } {dx}=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}+\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{7}+\frac{\mathrm{10}^{\mathrm{2}} }{\mathrm{32}+\frac{\mathrm{42}^{\mathrm{2}} }{\mathrm{174}+\frac{\mathrm{216}^{\mathrm{2}} }{\mathrm{1196}+\frac{\mathrm{1312}^{\mathrm{2}} }{....}}}}}}} \\$$

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

$$\int_{\mathrm{0}} ^{{a}} {e}^{−{x}^{\mathrm{2}} } {dx}=\int_{\mathrm{0}} ^{{a}} \mathrm{1}−{x}^{\mathrm{2}} +\frac{{x}^{\mathrm{4}} }{\mathrm{2}!}−\frac{{x}^{\mathrm{6}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{8}} }{\mathrm{4}!}−... \\$$$$={a}−\frac{{a}^{\mathrm{3}} }{\mathrm{3}.\mathrm{1}!}+\frac{{a}^{\mathrm{5}} }{\mathrm{5}.\mathrm{2}!}−\frac{{a}^{\mathrm{7}} }{\mathrm{7}.\mathrm{3}!}+\frac{{a}^{\mathrm{9}} }{\mathrm{9}.\mathrm{4}!}−.. \\$$$$={a}+{a}\left(−\frac{{a}^{\mathrm{2}} }{\mathrm{3}.\mathrm{1}!}\right)+{a}.\left(\frac{−{a}^{\mathrm{2}} }{\mathrm{3}.\mathrm{1}!}\right)\left(\frac{−\mathrm{3}.\mathrm{1}!{a}^{\mathrm{2}} }{\mathrm{5}.\mathrm{2}!}\right)+{a}\left(\frac{−{a}^{\mathrm{2}} }{\mathrm{3}.\mathrm{1}!}\right)\left(−\frac{\mathrm{3}.\mathrm{1}!{a}^{\mathrm{2}} }{\mathrm{5}.\mathrm{2}!}\right)\left(\frac{−\mathrm{5}.\mathrm{2}!}{\mathrm{7}.\mathrm{3}!}{a}^{\mathrm{2}} \right)+.. \\$$$$=\frac{{a}}{\mathrm{1}+\frac{\frac{{a}^{\mathrm{2}} }{\mathrm{3}.\mathrm{1}!}}{\mathrm{1}−\frac{{a}^{\mathrm{2}} }{\mathrm{3}.\mathrm{1}!}+\frac{\frac{\mathrm{3}.\mathrm{1}!{a}^{\mathrm{2}} }{\mathrm{5}.\mathrm{2}!}}{\mathrm{1}−\frac{\mathrm{3}.\mathrm{1}!{a}^{\mathrm{2}} }{\mathrm{5}.\mathrm{2}!}+\frac{\frac{\mathrm{5}.\mathrm{2}!{a}^{\mathrm{2}} }{\mathrm{7}.\mathrm{3}!}}{\mathrm{1}−\frac{\mathrm{5}.\mathrm{2}!}{\mathrm{7}.\mathrm{3}!}{a}^{\mathrm{2}} +...}}\:}} \\$$$$=\frac{{a}}{\mathrm{1}+\frac{{a}^{\mathrm{2}} }{\mathrm{3}.\mathrm{1}!−{a}^{\mathrm{2}} +\frac{\left(\mathrm{3}.\mathrm{1}!\right)^{\mathrm{2}} {a}^{\mathrm{2}} }{\mathrm{5}.\mathrm{2}!−\mathrm{3}.\mathrm{1}!{a}^{\mathrm{2}} +\frac{\left(\mathrm{5}.\mathrm{2}!\right)^{\mathrm{2}} {a}^{\mathrm{2}} }{\mathrm{7}.\mathrm{3}!−\mathrm{5}.\mathrm{2}!{a}^{\mathrm{2}} +\left(\mathrm{7}.\mathrm{3}!\right)^{\mathrm{2}} ...}}}} \\$$$$\\$$