Question Number 92156 by abdomathmax last updated on 05/May/20
$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{xe}^{−{x}^{\mathrm{2}} } {ln}\left(\mathrm{1}+{x}\right){dx} \\ $$
Commented by mathmax by abdo last updated on 08/May/20
![I =∫_0 ^1 x e^(−x^2 ) ln(1+x)dx by psrts I =[−(1/2)e^(−x^2 ) ln(1+x)]_0 ^1 +(1/2)∫_0 ^1 (e^(−x^2 ) /(1+x))dx =−(1/2)e^(−1) ln(2)+(1/2)∫_0 ^1 (e^(−x^2 ) /(1+x))dx ∫_0 ^1 (e^(−x^2 ) /(1+x))dx =_(1+x=t) ∫_1 ^2 (e^(−(t−1)^2 ) /t)dt =∫_1 ^2 (e^(−t^2 +2t −1) /t)dt =e^(−1) ∫_1 ^2 (e^(−t^2 +2t) /t) dt =e^(−1) ∫_1 ^2 (1/t)Σ_(n=0) ^∞ (((−t^2 +2t)^n )/(n!))dt =e^(−1) ∫_1 ^2 ((1/t)+(1/t)Σ_(n=1) ^∞ (((−t^2 +2t)^n )/(n!)))dt =e^(−1) ln(2) +e^(−1) Σ_(n=1) ^∞ ∫_1 ^2 ((t^(n−1) (2−t)^n )/(n!))dt ⇒ I = (1/(2e))Σ_(n=1) ^∞ (A_n /(n!)) with A_n =∫_1 ^2 t^(n−1) (2−t)^n dt ....be continued...](Q92715.png)
$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}\:{e}^{−{x}^{\mathrm{2}} } {ln}\left(\mathrm{1}+{x}\right){dx}\:\:{by}\:{psrts} \\ $$$${I}\:=\left[−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−{x}^{\mathrm{2}} } {ln}\left(\mathrm{1}+{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{e}^{−{x}^{\mathrm{2}} } }{\mathrm{1}+{x}}{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{1}} {ln}\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{e}^{−{x}^{\mathrm{2}} } }{\mathrm{1}+{x}}{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{e}^{−{x}^{\mathrm{2}} } }{\mathrm{1}+{x}}{dx}\:=_{\mathrm{1}+{x}={t}} \:\:\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{{e}^{−\left({t}−\mathrm{1}\right)^{\mathrm{2}} } }{{t}}{dt}\:=\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{{e}^{−{t}^{\mathrm{2}} +\mathrm{2}{t}\:−\mathrm{1}} }{{t}}{dt} \\ $$$$={e}^{−\mathrm{1}} \:\int_{\mathrm{1}} ^{\mathrm{2}} \:\frac{{e}^{−{t}^{\mathrm{2}} \:+\mathrm{2}{t}} }{{t}}\:{dt}\:={e}^{−\mathrm{1}} \:\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\mathrm{1}}{{t}}\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−{t}^{\mathrm{2}} +\mathrm{2}{t}\right)^{{n}} }{{n}!}{dt} \\ $$$$={e}^{−\mathrm{1}} \:\int_{\mathrm{1}} ^{\mathrm{2}} \left(\frac{\mathrm{1}}{{t}}+\frac{\mathrm{1}}{{t}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−{t}^{\mathrm{2}} +\mathrm{2}{t}\right)^{{n}} }{{n}!}\right){dt} \\ $$$$={e}^{−\mathrm{1}} {ln}\left(\mathrm{2}\right)\:+{e}^{−\mathrm{1}} \:\sum_{{n}=\mathrm{1}} ^{\infty} \:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\frac{{t}^{{n}−\mathrm{1}} \left(\mathrm{2}−{t}\right)^{{n}} }{{n}!}{dt}\:\:\Rightarrow \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{2}{e}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{A}_{{n}} }{{n}!}\:\:\:\:{with}\:\:{A}_{{n}} =\int_{\mathrm{1}} ^{\mathrm{2}} \:{t}^{{n}−\mathrm{1}} \left(\mathrm{2}−{t}\right)^{{n}} \:{dt}\:....{be}\:{continued}... \\ $$