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Question Number 151738 by tabata last updated on 22/Aug/21

$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \:{x}\:{d}\left({e}^{{x}^{\mathrm{2}} } \right) \\$$$$\\$$$${how}\:{it}\:{solve}\: \\$$

Answered by Olaf_Thorendsen last updated on 22/Aug/21

$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {xd}\left({e}^{{x}^{\mathrm{2}} } \right) \\$$$$\mathrm{I}\:=\:\left[{xe}^{{x}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\mathrm{1}} −\:\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{{x}^{\mathrm{2}} } {dx} \\$$$$\mathrm{I}\:=\:{e}−\frac{\sqrt{\pi}}{\mathrm{2}}\left[\mathrm{erfi}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\$$$$\mathrm{I}\:=\:{e}−\frac{\sqrt{\pi}}{\mathrm{2}}\mathrm{erfi}\left(\mathrm{1}\right) \\$$$$\mathrm{erfi}\left({z}\right)\:=\:\frac{\mathrm{2}}{\:\sqrt{\pi}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{z}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right){n}!} \\$$$$\mathrm{I}\:=\:{e}−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right){n}!} \\$$

Commented by Olaf_Thorendsen last updated on 22/Aug/21

$$\mathrm{I}\:\approx\:{e}−\mathrm{1},\mathrm{462651746} \\$$$$\mathrm{I}\:\approx\:\mathrm{1},\mathrm{255630082} \\$$