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let-give-x-0-t-x-1-e-t-dt-with-x-gt-0-prove-that-lim-n-gt-0-n-1-t-n-n-t-x-1-dt-x-

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Question Number 26566 by abdo imad last updated on 26/Dec/17
let give Γ(x)= ∫_0 ^∞ t^(x−1) e^(−t) dt with x>0 prove that lim _(n−>∝) ∫_0 ^n (1−(t/n))^n t^(x−1) dt = Γ(x)
letgiveΓ(x)=∫0∞tx−1e−tdtwithx>0provethatlimn−>∝∫0n(1−tn)ntx−1dt=Γ(x)
Commented by abdo imad last updated on 28/Dec/17
let put A_n = ∫_0 ^n (1−(t/n))^n t^(x−1) dt A_n = ∫_R f_n (t)dt where the sequence f_n = (1−(t/n))^n t^(x−1) χ_([0,n]) (t)we have f_n −>_(c.s) f(t)= e^(−t) t^(x−1) on [0,∝[ but /f_n /≤ ρ(t)= e^(−t ) t^(x−1 ) χ_([0,α) by theoreme of vonvergence domine ∫_R f_n (t)dt_(n−∝) −−>∫_0 ^∞ e^(−t) t^(x−1) dx=Γ(x)
letputAn=∫0n(1−tn)ntx−1dtAn=∫Rfn(t)dtwherethesequencefn=(1−tn)ntx−1χ[0,n](t)wehavefn−>c.sf(t)=e−ttx−1on[0,∝[but/fn/⩽ρ(t)=e−ttx−1χ[0,αbytheoremeofvonvergencedomine∫Rfn(t)dtn−∝−−>∫0∞e−ttx−1dx=Γ(x)

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