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
![I = ∫_0 ^1 xd(e^x^2 ) I = [xe^x^2 ]_0 ^1 − ∫_0 ^1 e^x^2 dx I = e−((√π)/2)[erfi(x)]_0 ^1 I = e−((√π)/2)erfi(1) erfi(z) = (2/( (√π)))Σ_(n=0) ^∞ (z^(2n+1) /((2n+1)n!)) I = e−Σ_(n=0) ^∞ (1/((2n+1)n!))](Q151740.png)
$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {xd}\left({e}^{{x}^{\mathrm{2}} } \right) \\ $$$$\mathrm{I}\:=\:\left[{xe}^{{x}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\mathrm{1}} −\:\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{{x}^{\mathrm{2}} } {dx} \\ $$$$\mathrm{I}\:=\:{e}−\frac{\sqrt{\pi}}{\mathrm{2}}\left[\mathrm{erfi}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\mathrm{I}\:=\:{e}−\frac{\sqrt{\pi}}{\mathrm{2}}\mathrm{erfi}\left(\mathrm{1}\right) \\ $$$$\mathrm{erfi}\left({z}\right)\:=\:\frac{\mathrm{2}}{\:\sqrt{\pi}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{z}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right){n}!} \\ $$$$\mathrm{I}\:=\:{e}−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right){n}!} \\ $$
Commented by Olaf_Thorendsen last updated on 22/Aug/21
$$\mathrm{I}\:\approx\:{e}−\mathrm{1},\mathrm{462651746} \\ $$$$\mathrm{I}\:\approx\:\mathrm{1},\mathrm{255630082} \\ $$