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Question Number 206794 by MaruMaru last updated on 25/Apr/24

help me... ∫_0 ^∞ ((sin(t)ln(t))/t)e^(−t) dt

helpme...∫0∞sin(t)ln(t)te−tdt

Answered by Berbere last updated on 25/Apr/24

let f(a)=∫_0 ^∞ (1/t)sin(at)ln(t)e^(−t) dt a∈R f is defined f(0)=0 We want f(1) we can verifie That f is C_1 ∀t∈]0,∞[ a→^g ((sin(at))/t)ln(t)e^(−t) is deivable ∀a∈R t→((sin(at))/t)ln(t)e^(−t) is integrabe g′(a)=cos(at)ln(t)e^(−t) ∣g′(a)∣≤∣ln(t)∣e^(−t) ;∀a∈R ∫_0 ^∞ ∣ln(t)∣e^(−t) dt<∞⇒we can switch ∂ and ∫ f′(a)=∫_0 ^∞ cos(at)ln(t)e^(−t) =Re∫_0 ^∞ ln(t)e^(t(−1+ia)) dt introduce h(z)=∫_0 ^∞ t^z e^(t(−1+ia)) ;z≥0 f′(a)=h′(z)∣_(z=0) h(z)=^((−1+ia)t→t) −∫_0 ^∞ ((t^z e^(−t) )/((−1+ia)^(z+1) ))dt=((Γ(1+z))/((−1+ia)^(z+1) )) f′(a)=h′(a)=−((Γ′(1)(−1+ia)−ln(−1+ia)(−1+ia))/((−1+ia)^2 )) =−(((Γ′(1))/(−1+ia))−((ln(−1+ia))/(−1+ia)));Γ′(1)=−γ∴Γ′(1)=Γ(1)Ψ(1)=1.−γ f′(a)=−Re((γ/(1−ia))+((ln(−1+ia))/(1−ia))) Re is lineair f(1)=∫_0 ^1 f′(a)da;=−Re∫_0 ^1 ((γ/(1−ia))+((ln(−1+ia))/(1−ia)))da =−Re[(γ/(−i))ln(1−ia)+(1/(−2i))ln^2 (−1+ia)]_0 ^1 =−Re[γiln(1−i)+(i/2)ln^2 (1−i)−(i/2)ln^2 (−1)] log(z)=ln∣z∣ +i arg(z);z∈]−π,π] =−Re(γi(ln((√2))−((iπ)/4)+(i/2)(ln(√2)−((iπ)/4))^2 −(i/2)(iπ)^2 ) =−(((γπ)/4)+(π/4)ln((√2)))=−(π/4)γ−((πln(2))/8) ∫_0 ^∞ ((sin(t)ln(t)e^(−t) )/t)dt=−(π/4)(γ+((ln(2))/2))

letf(a)=∫0∞1tsin(at)ln(t)e−tdta∈Rfisdefinedf(0)=0Wewantf(1)wecanverifieThatfisC1∀t∈]0,∞[a→gsin(at)tln(t)e−tisdeivable∀a∈Rt→sin(at)tln(t)e−tisintegrabeg′(a)=cos(at)ln(t)e−t∣g′(a)∣⩽∣ln(t)∣e−t;∀a∈R∫0∞∣ln(t)∣e−tdt<∞⇒wecanswitch∂and∫f′(a)=∫0∞cos(at)ln(t)e−t=Re∫0∞ln(t)et(−1+ia)dtintroduceh(z)=∫0∞tzet(−1+ia);z⩾0f′(a)=h′(z)∣z=0h(z)=(−1+ia)t→t−∫0∞tze−t(−1+ia)z+1dt=Γ(1+z)(−1+ia)z+1f′(a)=h′(a)=−Γ′(1)(−1+ia)−ln(−1+ia)(−1+ia)(−1+ia)2=−(Γ′(1)−1+ia−ln(−1+ia)−1+ia);Γ′(1)=−γ∴Γ′(1)=Γ(1)Ψ(1)=1.−γf′(a)=−Re(γ1−ia+ln(−1+ia)1−ia)Reislineairf(1)=∫01f′(a)da;=−Re∫01(γ1−ia+ln(−1+ia)1−ia)da=−Re[γ−iln(1−ia)+1−2iln2(−1+ia)]01=−Re[γiln(1−i)+i2ln2(1−i)−i2ln2(−1)]log(z)=ln∣z∣+iarg(z);z∈]−π,π]=−Re(γi(ln(2)−iπ4+i2(ln2−iπ4)2−i2(iπ)2)=−(γπ4+π4ln(2))=−π4γ−πln(2)8∫0∞sin(t)ln(t)e−ttdt=−π4(γ+ln(2)2)

Commented by MaruMaru last updated on 26/Apr/24

°⌣° good thx!

°⌣°goodthx!

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