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let-f-x-0-e-t-1-xt-dt-calculate-f-n-0-

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Question Number 32739 by caravan msup abdo. last updated on 01/Apr/18
let f(x)=∫_0 ^∞ (e^(−t) /(1+xt))dt calculate f^((n)) (0).
letf(x)=∫0∞e−t1+xtdtcalculatef(n)(0).
Commented by abdo imad last updated on 03/Apr/18
f is C^∞ and f^′ (x) =∫_0 ^∞ (∂/∂x)( (e^(−t) /(1+xt)))dt = ∫_0 ^∞ ((−t e^(−t) )/((1+xt)^2 )) dt also we have f(x) =∫_0 ^∞ (e^(−t) /t) (1/(x+(1/t)))dt ⇒ f^((n)) (x) = ∫_0 ^∞ (e^(−t) /t) (((−1)^n n!)/((x+(1/t))^(n+1) ))dt = (−1)^n n! ∫_0 ^∞ (e^(−t) /t) (t^(n+1) /((1+xt)^(n+1) )) dt =(−1)^n (n!) ∫_0 ^∞ ((t^n e^(−t) )/((1+xt)^(n+1) ))dt ⇒ f^((n)) (0) = (−1)^n (n!) ∫_0 ^∞ t^n e^(−t) dt let calculate A_(n ) =∫_0 ^∞ t^n e^(−t) dt .by parts A_n = [−t^n e^(−t) ]_0 ^∞ +∫_0 ^∞ n t^(n−1) e^(−t) dt =n A_(n−1) ⇒ Π_(k=1) ^n A_k =n! Π_(k=1) ^n A_(k−1) ⇒ A_n =n! A_0 =n! (look also that A_n =Γ(n+1) =n!) ⇒f^((n)) (0) =(−1)^n (n!)^2 .
fisC∞andf′(x)=∫0∞∂∂x(e−t1+xt)dt=∫0∞−te−t(1+xt)2dtalsowehavef(x)=∫0∞e−tt1x+1tdt⇒f(n)(x)=∫0∞e−tt(−1)nn!(x+1t)n+1dt=(−1)nn!∫0∞e−tttn+1(1+xt)n+1dt=(−1)n(n!)∫0∞tne−t(1+xt)n+1dt⇒f(n)(0)=(−1)n(n!)∫0∞tne−tdtletcalculateAn=∫0∞tne−tdt.bypartsAn=[−tne−t]0∞+∫0∞ntn−1e−tdt=nAn−1⇒∏k=1nAk=n!∏k=1nAk−1⇒An=n!A0=n!(lookalsothatAn=Γ(n+1)=n!)⇒f(n)(0)=(−1)n(n!)2.

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