Question Number 73396 by mathmax by abdo last updated on 11/Nov/19
$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{t}^{\mathrm{2}} } {ln}\left(\mathrm{1}−{t}\right){dt} \\ $$
Commented by mathmax by abdo last updated on 14/Nov/19
$${let}\:{take}\:{a}\:{try}\:\:{let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{t}^{\mathrm{2}} } {ln}\left(\mathrm{1}−{t}\right){dt}\:\:{we}\:{have} \\ $$$${ln}^{'} \left(\mathrm{1}−{t}\right)\:=\frac{−\mathrm{1}}{\mathrm{1}−{t}}\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:{t}^{{n}} \:\:\:{for}\:\mid{t}\mid<\mathrm{1}\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}−{t}\right)\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{t}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:\:\:+{c}\:\:\:\:\left({c}=\mathrm{0}\right) \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{t}^{{n}} }{{n}}\:\Rightarrow{I}\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{−{t}^{\mathrm{2}} } \left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{t}^{{n}} }{{n}}\right){dt} \\ $$$$=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{t}^{{n}} {e}^{−{t}^{\mathrm{2}} \:} \:\:{dt}\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{W}_{{n}} }{{n}}\:\:{with} \\ $$$${W}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{{n}} \:{e}^{−{t}^{\mathrm{2}} } \:{dt}\:\:\:{by}\:{parts}\:{u}^{'} ={t}^{{n}} \:{and}\:{v}={e}^{−{t}^{\mathrm{2}} } \:\Rightarrow \\ $$$${W}_{{n}} =\left[\frac{\mathrm{1}}{{n}+\mathrm{1}}{t}^{{n}+\mathrm{1}} \:{e}^{−{t}^{\mathrm{2}} } \right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{{n}+\mathrm{1}}{t}^{{n}+\mathrm{1}} \left(−\mathrm{2}{t}\right){e}^{−{t}^{\mathrm{2}} } {dt} \\ $$$$=\frac{{e}^{−\mathrm{1}} }{{n}+\mathrm{1}}\:+\frac{\mathrm{2}}{{n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{t}^{{n}+\mathrm{2}} \:{e}^{−{t}^{\mathrm{2}} } {dt}\:=\frac{{e}^{−\mathrm{1}} }{{n}+\mathrm{1}}\:+\frac{\mathrm{2}}{{n}+\mathrm{1}}\:{W}_{{n}+\mathrm{2}} \:\Rightarrow \\ $$$$\left({n}+\mathrm{1}\right){W}_{{n}} ={e}^{−\mathrm{1}} \:+\mathrm{2}\:{W}_{{n}+\mathrm{2}} \:\Rightarrow{W}_{{n}+\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\left({n}+\mathrm{1}\right){W}_{{n}} −{e}^{−\mathrm{1}} \right\} \\ $$$$….{be}\:{continued}…. \\ $$
Answered by mind is power last updated on 11/Nov/19
$${this}\:{a}\:{tricky}\:{one}\:{We}\:{use}\:{F}_{\mathrm{2}} \left({a},{b},{c},{z}\right)\:\:{hypdrgeometric}\:{function} \\ $$