Question Number 37281 by abdo.msup.com last updated on 11/Jun/18
$${find}\:{a}\:{better}\:{approximation}\:{for}\:{the} \\ $$$${integrals}\: \\ $$$$\left.\mathrm{1}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{e}^{−{x}^{\mathrm{2}} } {dx} \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{1}} ^{+\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:. \\ $$
Commented by prof Abdo imad last updated on 18/Jun/18
$${we}\:{have}\:\:{e}^{−{x}^{\mathrm{2}} } \:\:\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \:{x}^{\mathrm{2}{n}} }{{n}!}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{\mathrm{2}{n}} {dx} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!\left(\mathrm{2}{n}+\mathrm{1}\right)}\:{let}\:{take}\:\mathrm{8}{terms}\:{we}\:{get} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:\sim\:\mathrm{1}\:−\frac{\mathrm{1}}{\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{5}.\mathrm{2}!}\:\:−\frac{\mathrm{1}}{\mathrm{7}.\mathrm{3}!}\:+\frac{\mathrm{1}}{\mathrm{9}.\mathrm{4}!} \\ $$$$−\frac{\mathrm{1}}{\mathrm{11}.\mathrm{5}!}\:\:+\frac{\mathrm{1}}{\mathrm{13}.\mathrm{6}!}\:−\frac{\mathrm{1}}{\mathrm{15}.\mathrm{7}!}\:+\frac{\mathrm{1}}{\mathrm{17}.\mathrm{8}!}\:\:\Rightarrow… \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{+\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:+\int_{\mathrm{1}} ^{+\infty} \:{e}^{−{x}^{\mathrm{2}} } {dx} \\ $$$$\Rightarrow\:\int_{\mathrm{1}} ^{+\infty} \:\:{e}^{−{x}^{\mathrm{2}} } {dx}\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{x}^{\mathrm{2}} } {dx}\:{and}\:{we}\:{use} \\ $$$${approximate}\:{value}\:{from}\:{Q}\mathrm{1}… \\ $$