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let-f-x-0-t-a-1-x-t-n-dt-with-0-lt-a-lt-1-and-x-gt-0-and-n-2-1-determine-a-explicit-form-of-f-x-2-calculate-g-x-0-t-a-1-x-t-n-2-dt-3-find-f-k-x-at-for

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Question Number 63664 by mathmax by abdo last updated on 07/Jul/19
let f(x)=∫_0 ^∞ (t^(a−1) /(x+t^n )) dt with 0<a<1 and x>0 and n≥2 1) determine a explicit form of f(x) 2) calculate g(x) =∫_0 ^∞ (t^(a−1) /((x+t^n )^2 )) dt 3) find f^((k)) (x) at form of integrals 4) calculate ∫_0 ^∞ (t^(a−1) /(9+t^2 )) dt and ∫_0 ^∞ (t^(a−1) /((9+t^2 )^2 )) 5) calculate U_n =∫_0 ^∞ (t^((1/n)−1) /(2^n +t^n )) dt and study the convergence of Σ U_n
letf(x)=∫0∞ta−1x+tndtwith0<a<1andx>0andn⩾21)determineaexplicitformoff(x)2)calculateg(x)=∫0∞ta−1(x+tn)2dt3)findf(k)(x)atformofintegrals4)calculate∫0∞ta−19+t2dtand∫0∞ta−1(9+t2)25)calculateUn=∫0∞t1n−12n+tndtandstudytheconvergenceofΣUn
Commented by mathmax by abdo last updated on 09/Jul/19
1)we have f(x)=∫_0 ^∞ (t^(a−1) /(x +t^n ))dt changement t =^n (√x)u give f(x) =∫_0 ^∞ (((^n (√x)u)^(a−1) )/(x +xu^n ))^n (√x)du =(x^(((a−1)/n)+(1/n)) /x) ∫_0 ^∞ (u^(a−1) /(1+u^n )) du =x^((a/n)−1) ∫_0 ^∞ (u^(a−1) /(1+u^n )) =_(u=α^(1/n) ) x^((a/n)−1) ∫_0 ^∞ (α^((a−1)/n) /(1+α)) (1/n) α^((1/n)−1) dα = x^((a/n)−1) ∫_0 ^∞ (α^(((a−1)/n)+(1/n)−1) /(1+α)) dα = x^((a/n)−1) ∫_0 ^∞ (α^((a/n)−1) /(1+α)) dα ⇒ f(x)=x^((a/n)−1) (π/(sin(((πa)/n))))
1)wehavef(x)=∫0∞ta−1x+tndtchangementt=nxugivef(x)=∫0∞(nxu)a−1x+xunnxdu=xa−1n+1nx∫0∞ua−11+undu=xan−1∫0∞ua−11+un=u=α1nxan−1∫0∞αa−1n1+α1nα1n−1dα=xan−1∫0∞αa−1n+1n−11+αdα=xan−1∫0∞αan−11+αdα⇒f(x)=xan−1πsin(πan)
Commented by mathmax by abdo last updated on 09/Jul/19
2) we have by derivation f^′ (x) =∫_0 ^∞ (∂/∂x)((t^(a−1) /(x+t^n )))dt =−∫_0 ^∞ (t^(a−1) /((x+t^n )^2 ))dt =−g(x) ⇒g(x)=−f^′ (x) f(x)=(π/(sin(((πa)/n)))) x^((a/n)−1) ⇒f^′ (x) =((π((a/n)−1))/(sin(((πa)/n)))) x^((a/n)−2) ⇒ g(x)=∫_0 ^∞ (t^(a−1) /((x+t^n )^2 ))dt =((π(1−(a/n)))/(sin(((πa)/n)))) x^((a/n)−2)
2)wehavebyderivationf′(x)=∫0∞∂∂x(ta−1x+tn)dt=−∫0∞ta−1(x+tn)2dt=−g(x)⇒g(x)=−f′(x)f(x)=πsin(πan)xan−1⇒f′(x)=π(an−1)sin(πan)xan−2⇒g(x)=∫0∞ta−1(x+tn)2dt=π(1−an)sin(πan)xan−2
Commented by mathmax by abdo last updated on 09/Jul/19
3) we have f(x)=∫_0 ^∞ (t^(a−1) /(x +t^n )) dt ⇒f^((k)) (x) =∫_0 ^∞ (∂^k /∂x^k )((t^(a−1) /(x+t^n )))dt =∫_0 ^∞ (((−1)^k k!)/((x+t^n )^(k+1) )) t^(a−1) dt ⇒f^((k)) (x) =(−1)^k k! ∫_0 ^∞ (t^(a−1) /((x+t^n )^(k+1) ))dt
3)wehavef(x)=∫0∞ta−1x+tndt⇒f(k)(x)=∫0∞∂k∂xk(ta−1x+tn)dt=∫0∞(−1)kk!(x+tn)k+1ta−1dt⇒f(k)(x)=(−1)kk!∫0∞ta−1(x+tn)k+1dt
Commented by mathmax by abdo last updated on 09/Jul/19
4) we have proved that ∫_0 ^∞ (t^(a−1) /(x+t^n )) dt =(π/(sin(((πa)/n)))) x^((a/n)−1) and ∫_0 ^∞ (t^(a−1) /((x+t^n )^2 ))dt =((π(1−(a/n)))/(sin(((πa)/n)))) x^((a/n)−2) x=9 and n=2 ⇒∫_0 ^∞ (t^(a−1) /(9+t^2 ))dt =(π/(sin(((πa)/2))))×9^((a/2)−1) ∫_0 ^∞ (t^(a−1) /((9+t^2 )^2 ))dt =((π(1−(a/2)))/(sin(((πa)/2))))×9^((a/2)−2)
4)wehaveprovedthat∫0∞ta−1x+tndt=πsin(πan)xan−1and∫0∞ta−1(x+tn)2dt=π(1−an)sin(πan)xan−2x=9andn=2⇒∫0∞ta−19+t2dt=πsin(πa2)×9a2−1∫0∞ta−1(9+t2)2dt=π(1−a2)sin(πa2)×9a2−2
Commented by mathmax by abdo last updated on 09/Jul/19
5) we have ∫_0 ^∞ (t^(a−1) /(x+t^n ))dt =(π/(sin(((πa)/n)))) x^((a/n)−1) a=(1/n) and x =2^n ⇒∫_0 ^∞ (t^((1/n)−1) /(2^n +t^n ))dt = (π/(sin((π/n^2 ))))(2^n )^((1/n^2 )−1) =(π/(sin((π/n^2 ))))×2^((1/n)−n) =(π/(2^(n−(1/n)) sin((π/n^2 )))) .
5)wehave∫0∞ta−1x+tndt=πsin(πan)xan−1a=1nandx=2n⇒∫0∞t1n−12n+tndt=πsin(πn2)(2n)1n2−1=πsin(πn2)×21n−n=π2n−1nsin(πn2).
Commented by mathmax by abdo last updated on 09/Jul/19
⇒ U_n =(π/(2^(n−(1/n)) sin((π/n^2 ))))
⇒Un=π2n−1nsin(πn2)

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