Integration Questions

Question Number 100047 by Ar Brandon last updated on 24/Jun/20

$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{dx} \\$$

Answered by smridha last updated on 24/Jun/20

$$=\frac{\sqrt{\boldsymbol{\pi}}}{\mathrm{2}}{erf}\left(\mathrm{1}\right) \\$$

Commented by Ar Brandon last updated on 24/Jun/20

Thanks, but how do we arrive there ? Or is it just a theory ?

Commented by smridha last updated on 24/Jun/20

$$\boldsymbol{{yeah}}\:\boldsymbol{{you}}\:\boldsymbol{{can}}\:\boldsymbol{{say}}\:\boldsymbol{{that}} \\$$$$\boldsymbol{{the}}\:\boldsymbol{{fact}}\:\boldsymbol{{is}}.. \\$$$$\boldsymbol{{erf}}\left(\boldsymbol{{x}}\right)=\frac{\mathrm{2}}{\sqrt{\boldsymbol{\pi}}}\int_{\mathrm{0}} ^{{x}} \boldsymbol{{e}}^{−\boldsymbol{{t}}^{\mathrm{2}} } \boldsymbol{{dt}} \\$$$$\boldsymbol{{well}}\:\boldsymbol{{I}}\:\boldsymbol{{have}}\:\boldsymbol{{defferent}}\:\boldsymbol{{way}}\:\boldsymbol{{wait}} \\$$$$\boldsymbol{{I}}\:\boldsymbol{{post}}\:\boldsymbol{{it}}... \\$$

Answered by smridha last updated on 24/Jun/20

$$=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \boldsymbol{{x}}^{\mathrm{2}\boldsymbol{{n}}} }{\boldsymbol{{n}}!}\boldsymbol{{dx}} \\$$$$=\mathrm{1}.\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} }{\boldsymbol{{n}}!\left(\mathrm{2}\boldsymbol{{n}}+\mathrm{1}\right)} \\$$$$=\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\left(\boldsymbol{{n}}+\frac{\mathrm{1}}{\mathrm{2}}\right)}.\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} }{\boldsymbol{{n}}!} \\$$$$=\underset{\boldsymbol{{n}}=\mathrm{0}\:} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{\boldsymbol{{n}}} }{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)_{\boldsymbol{{n}}} }.\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{n}}} }{\boldsymbol{{n}}!} \\$$$$=\boldsymbol{{M}}\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{2}},−\mathrm{1}\right)\boldsymbol{{or}}\:_{\mathrm{1}} \boldsymbol{{F}}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}};\frac{\mathrm{3}}{\mathrm{2}};−\mathrm{1}\right) \\$$$$\boldsymbol{{this}}\:\boldsymbol{{is}}\:\boldsymbol{{called}}\:\boldsymbol{{confluent}}\:\boldsymbol{{hypergeometric}} \\$$$$\boldsymbol{{f}}^{\boldsymbol{{n}}} . \\$$$$\boldsymbol{{note}}:\left(\boldsymbol{{a}}\right)=\frac{\left(\boldsymbol{{a}}\right)_{\boldsymbol{{n}}} }{\left(\boldsymbol{{a}}+\mathrm{1}\right)_{\boldsymbol{{n}}} }\left(\boldsymbol{{a}}+\boldsymbol{{n}}\right)\left[\boldsymbol{{pochhammer}}\:\boldsymbol{{symbol}}\right] \\$$

Commented by Ar Brandon last updated on 24/Jun/20

Thanks so very much��