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Question-83558




Question Number 83558 by Jidda28 last updated on 03/Mar/20
Commented by Jidda28 last updated on 03/Mar/20
help me out pls.
$${help}\:{me}\:{out}\:{pls}. \\ $$
Commented by mathmax by abdo last updated on 03/Mar/20
A_n =∫_0 ^1  x^n (lnx)^n  dx   changement lnx=−t give  A_n =∫_(+∞) ^0  (e^(−t) )^n (−t)^n  (−e^(−t) )dt =∫_0 ^∞  e^(−(n+1)t)  t^n  dt  =_((n+1)t=u)    ∫_0 ^∞   e^(−u)  ((u/(n+1)))^n  (du/(n+1)) =(1/((n+1)^(n+1) )) ∫_0 ^∞  u^n  e^(−u)  du  we have Γ(x)=∫_0 ^∞  t^(x−1)  e^(−t)  dt   (x>0) ⇒A_n =(1/((n+1)^(n+1) ))×Γ(n+1)  =((n!)/((n+1)^(n+1) ))
$${A}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}} \left({lnx}\right)^{{n}} \:{dx}\:\:\:{changement}\:{lnx}=−{t}\:{give} \\ $$$${A}_{{n}} =\int_{+\infty} ^{\mathrm{0}} \:\left({e}^{−{t}} \right)^{{n}} \left(−{t}\right)^{{n}} \:\left(−{e}^{−{t}} \right){dt}\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left({n}+\mathrm{1}\right){t}} \:{t}^{{n}} \:{dt} \\ $$$$=_{\left({n}+\mathrm{1}\right){t}={u}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{u}} \:\left(\frac{{u}}{{n}+\mathrm{1}}\right)^{{n}} \:\frac{{du}}{{n}+\mathrm{1}}\:=\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:\int_{\mathrm{0}} ^{\infty} \:{u}^{{n}} \:{e}^{−{u}} \:{du} \\ $$$${we}\:{have}\:\Gamma\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:{t}^{{x}−\mathrm{1}} \:{e}^{−{t}} \:{dt}\:\:\:\left({x}>\mathrm{0}\right)\:\Rightarrow{A}_{{n}} =\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }×\Gamma\left({n}+\mathrm{1}\right) \\ $$$$=\frac{{n}!}{\left({n}+\mathrm{1}\right)^{{n}+\mathrm{1}} } \\ $$
Commented by Jidda28 last updated on 03/Mar/20
thank you very much sir, but x^(m  ^ )
$${thank}\:{you}\:{very}\:{much}\:{sir},\:{but}\:{x}^{{m}\:\overset{} {\:}} \\ $$
Commented by mathmax by abdo last updated on 03/Mar/20
you are welcome
$${you}\:{are}\:{welcome} \\ $$

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