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Question Number 119570 by mnjuly1970 last updated on 25/Oct/20
... ♣_♣ ^♣ nice calculus♣_♣ ^♣ ... prove that :: Ω=∫_0 ^( ∞) e^(−2x) ln(((1+e^(−x) )/(1−e^(−x) )))=1 ...★ M.N.1970★...
…♣♣♣nicecalculus♣♣♣…provethat::Ω=∫0∞e−2xln(1+e−x1−e−x)=1…★M.N.1970★…
Answered by mindispower last updated on 25/Oct/20
e^(−2x) =e^(−x) ∗e^(−x) let t=e^(−x) ⇒dt=−e^(−x) dx ⇔=∫_0 ^1 tln(((1+t)/(1−t)))dt=−∫_0 ^1 tln(((1−t)/(1+t)))dt x=((1−t)/(1+t))⇒t=((1−x)/(1+x)),dt=((−2)/((1+x)^2 ))dx ⇔−2∫_0 ^1 ((1−x)/((1+x)^3 ))ln(x)dx =−2[((1/((1+x)))−(1/((1+x)^2 )))ln(x)]_0 ^1 +2∫((1/((1+x)x))−(1/(x(1+x)^2 )))dx =2∫_0 ^1 (1/((1+x)^2 ))=2[−(1/((1+x)))]_0 ^1 =2.(1/2)=1
e−2x=e−x∗e−xlett=e−x⇒dt=−e−xdx⇔=∫01tln(1+t1−t)dt=−∫01tln(1−t1+t)dtx=1−t1+t⇒t=1−x1+x,dt=−2(1+x)2dx⇔−2∫011−x(1+x)3ln(x)dx=−2[(1(1+x)−1(1+x)2)ln(x)]01+2∫(1(1+x)x−1x(1+x)2)dx=2∫011(1+x)2=2[−1(1+x)]01=2.12=1
Commented by mnjuly1970 last updated on 25/Oct/20
thank you mr power..
thankyoumrpower..
Answered by mathmax by abdo last updated on 25/Oct/20
I =∫_0 ^∞ e^(−2x) ln(((1+e^(−x) )/(1−e^(−x) )))dx ⇒I =∫_0 ^∞ e^(−2x) ln(1+e^(−x) )dx−∫_0 ^∞ e^(−2x) ln(1−e^(−x) )dx =H−K H =∫_0 ^∞ e^(−2x) ln(1+e^(−x) )dx =_(e^(−x) =t) −∫_0 ^1 t^2 ln(1+t)(((−dt)/t))dt =∫_0 ^1 tln(1+t)dt =[(t^2 /2)ln(1+t)]_0 ^1 −∫_0 ^1 (t^2 /(2(1+t)))dt =((ln2)/2)−(1/2)∫_0 ^1 ((t^2 −1+1)/(t+1))dt =((ln2)/2)−(1/2){∫_0 ^1 (t−1)dt+∫_0 ^1 (dt/(t+1))} =((ln2)/2)−(1/2){[(t^2 /2)−t]_0 ^1 +ln(2)} =−(1/2)(−(1/2))=(1/4) K=∫_0 ^∞ e^(−2x) ln(1−e^(−x) )dx =_(e^(−x) =t) −∫_0 ^1 t^2 ln(1−t)(((−dt)/t)) =∫_0 ^1 tln(1−t)dt =[(1/2)(t^2 −1)ln(1−t)]_0 ^1 −∫_0 ^1 ((t^2 −1)/2)×((−1)/(1−t))dt =−(1/2)∫_0 ^1 (t+1)dt =−(1/2)[(t^2 /2)+t]_0 ^1 =−(1/2)((3/2))=−(3/4) ⇒ I =H−K =(1/4)+(3/4) ⇒ ★ I =1★
I=∫0∞e−2xln(1+e−x1−e−x)dx⇒I=∫0∞e−2xln(1+e−x)dx−∫0∞e−2xln(1−e−x)dx=H−KH=∫0∞e−2xln(1+e−x)dx=e−x=t−∫01t2ln(1+t)(−dtt)dt=∫01tln(1+t)dt=[t22ln(1+t)]01−∫01t22(1+t)dt=ln22−12∫01t2−1+1t+1dt=ln22−12{∫01(t−1)dt+∫01dtt+1}=ln22−12{[t22−t]01+ln(2)}=−12(−12)=14K=∫0∞e−2xln(1−e−x)dx=e−x=t−∫01t2ln(1−t)(−dtt)=∫01tln(1−t)dt=[12(t2−1)ln(1−t)]01−∫01t2−12×−11−tdt=−12∫01(t+1)dt=−12[t22+t]01=−12(32)=−34⇒I=H−K=14+34⇒★I=1★
Commented by mnjuly1970 last updated on 25/Oct/20
thank you mr max..
thankyoumrmax..
Commented by Bird last updated on 25/Oct/20
you are welcome
youarewelcome
Answered by mnjuly1970 last updated on 25/Oct/20
Commented by mnjuly1970 last updated on 25/Oct/20
edittion : 1: recall :: ln(((1+z)/(1−z)))=2Σ_(n=0) ^∞ (z^(2n+1) /(2n+1)) 2 : Ω=^(telescopic series) (1/(2(0)+1))−lim_(n→∞) ((1/(2n+3)))=1 ...m.n.july.1970...
edittion:1:recall::ln(1+z1−z)=2∑∞n=0z2n+12n+12:Ω=telescopicseries12(0)+1−limn→∞(12n+3)=1…m.n.july.1970…

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