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Question Number 66444 by ~ À ® @ 237 ~ last updated on 15/Aug/19
calculatelimx→0(x!)1xifx!=Π(x)=∫0∞txe−tdt
Commented by ~ À ® @ 237 ~ last updated on 18/Aug/19
letnameditLweknowthatL>0Soln(L)=limx→0ln[(x!)1x]=limx→0ln(x!)xaslimx→0ln(x!)=ln(Π(0))=ln(Γ(1))=ln(1)=0,wecanusetheHospitalRuleThenln(L)=limx→0∏′(x)Π(x)1=limx→0∫0∞txlnte−tdt∫0∞txe−tdt=∫0∞e−tlntdt∫0∞e−tdt=∫0∞e−tlntdtbecause∫0∞e−tdt=limx→∞(1−e−x)=1Nowwehaveln(L)=∫0∞e−tlntdt=∫0∞limn→∞(1−tn)nlntdt=limn→∞JnwithJn=∫0n(1−tn)nlntdtletexplicitJnintermsofnForthat,letnamedu=1−tn⇔t=n(1−u)⇒dt=−nduJn=n∫01unln(n(1−u))du=nln(n)∫01undu+n∫01unln(1−u)du=nln(n)n+1+nKnInthewaytofindKn,weascertainthatunln(1−u)+un+1−1(n+1)(u−1)=ddu[un+1−1n+1ln(1−u)]SowehaveKn=[un+1−1n+1ln(1−u)]01−1n+1∫01un+1−1u−1du=−1n+1∫01(1+u+u2+....+un)ducauselimu→1(un+1−1)ln(1−u)=limu→1−(1+u+....+un)(1−u)ln(1−u)=0(whenknowingthatlimx→0xlnx=0)=−1n+1∑nk=0∫01ukdu=−1n+1∑nk=01k+1=−1n+1∑n+1k=11kThenwereachatJn=nn+1ln(n)−nn+1∑n+1k=11k=nn+1[ln(n)−Hn+1]withHn=∑nk=11kwehaveHn+1=Hn+1n+1UsingthatwegetJn=nn+1[ln(n)−Hn−1n+1]=−n(n+1)2−nn+1[Hn−ln(n)]Soln(L)=limn→∞Jn=limn→∞−n(n+1)2−limn→∞nn+1.limn→∞(Hn−ln(n))=−γ(TheEuler′sconstant)FinallyL=limx→0(x!)1x=e−γ.
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