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Question Number 45045 by maxmathsup by imad last updated on 07/Oct/18
find the value of  ∫_0 ^∞    (t^3 /(1+e^t ))dt .
findthevalueof0t31+etdt.
Answered by maxmathsup by imad last updated on 09/Oct/18
let A =∫_0 ^∞  (t^3 /(1+e^t )) dt  we have A =∫_0 ^∞  ((t^3 e^(−t) )/(1+e^(−t) ))dt  =∫_0 ^∞  t^3 e^(−t) (Σ_(n=0) ^∞ (−1)^n e^(−nt) )dt=Σ_(n=0) ^∞ (−1)^n  ∫_0 ^∞  t^3  e^(−(n+1)t) dt  changement (n+1)t =x give   A =Σ_(n=0) ^∞  (−1)^n  ∫_0 ^∞ ((x/(n+1)))^3 e^(−x)   (dx/((n+1)))  =Σ_(n=0) ^∞  (((−1)^n )/((n+1)^4 )) ∫_0 ^∞   x^3  e^(−x) dt  let remember that Γ(x)=∫_0 ^∞  t^(x−1) e^(−t) dt (x>0) ⇒  ∫_0 ^∞  x^3  e^(−x)  dx =∫_0 ^∞ x^(4−1) e^(−x) dx =Γ(4) =(4−1)!=3! =6 ⇒  A =6 Σ_(n=0) ^∞   (((−1)^n )/((n+1)^4 )) =6 δ(4) with δ(x)=Σ_(n=1) ^∞  (((−1)^(n−1) )/n^x )   (x>0) let find δ(x)  interms of ξ(x)=Σ_(n=1) ^∞   (1/n^x )  we have δ(x)=−Σ_(n=1) ^∞   (1/((2n)^x )) +Σ_(n=0) ^∞   (1/((2n+1)^x ))  =−2^(−x)  ξ(x) +Σ_(n=0) ^∞   (1/((2n+1)^x )) but ξ(x)=Σ_(n=1) ^∞  (1/((2n)^x )) +Σ_(n=0) ^∞  (1/((2n+1)^x ))  =2^(−x) ξ(x)+Σ_(n=0) ^∞  (1/((2n+1)^x )) ⇒Σ_(n=0) ^∞  (1/((2n+1)^x )) =(1−2^(−x) )ξ(x) ⇒  δ(x) =−2^(−x) ξ(x) +(1−2^(−x) )ξ(x) =(1−2^(1−x) )ξ(x) ⇒  δ(4) =(1−2^(−3) )ξ(4) =(1−(1/8))(π^4 /(90)) =(7/8) (π^4 /(90)) ⇒A =6.(7/8) .(π^4 /(90)) ⇒  =((2.3.7)/(2.4)) .(π^4 /(3.30)) = ((7π^4 )/(120))⇒∫_0 ^∞   (t^3 /(1+e^t )) dt =((7 π^4 )/(120)) ★
letA=0t31+etdtwehaveA=0t3et1+etdt=0t3et(n=0(1)nent)dt=n=0(1)n0t3e(n+1)tdtchangement(n+1)t=xgiveA=n=0(1)n0(xn+1)3exdx(n+1)=n=0(1)n(n+1)40x3exdtletrememberthatΓ(x)=0tx1etdt(x>0)0x3exdx=0x41exdx=Γ(4)=(41)!=3!=6A=6n=0(1)n(n+1)4=6δ(4)withδ(x)=n=1(1)n1nx(x>0)letfindδ(x)intermsofξ(x)=n=11nxwehaveδ(x)=n=11(2n)x+n=01(2n+1)x=2xξ(x)+n=01(2n+1)xbutξ(x)=n=11(2n)x+n=01(2n+1)x=2xξ(x)+n=01(2n+1)xn=01(2n+1)x=(12x)ξ(x)δ(x)=2xξ(x)+(12x)ξ(x)=(121x)ξ(x)δ(4)=(123)ξ(4)=(118)π490=78π490A=6.78.π490=2.3.72.4.π43.30=7π41200t31+etdt=7π4120

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