Question Number 59503 by ANTARES VY last updated on 11/May/19
$$\underset{\boldsymbol{{a}}} {\overset{\boldsymbol{{b}}} {\int}}\left(\boldsymbol{{e}}^{−\boldsymbol{{x}}^{\mathrm{2}} } \right)\boldsymbol{{dx}}=? \\ $$
Commented by Mr X pcx last updated on 11/May/19
$${at}\:{form}\:{of}\:{serie}\:\:{we}\:{have}\:{e}^{−{x}^{\mathrm{2}} } =\sum_{{m}=\mathrm{0}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}} }{{n}!}\:\Rightarrow \\ $$$$\int_{{a}} ^{{b}} \:\:{e}^{−{x}^{\mathrm{2}} } {dx}\:=\sum_{{m}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\int_{{a}} ^{{b}} \:{x}^{\mathrm{2}{n}} {dx} \\ $$$$=\sum_{{m}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}\left\{{b}^{\mathrm{2}{n}+\mathrm{1}} −{a}^{\mathrm{2}{n}+\mathrm{1}} \right\} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right){n}!}\:{b}^{\mathrm{2}{n}+\mathrm{1}} −\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right){n}!}{a}^{\mathrm{2}{n}+\mathrm{1}} \\ $$