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Question Number 123674 by mathmax by abdo last updated on 27/Nov/20

find ∫_0 ^∞ e^(−2x) ln(1+e^x )dx

find∫0∞e−2xln(1+ex)dx

Answered by Dwaipayan Shikari last updated on 27/Nov/20

∫_0 ^∞ e^(−2x) log(1+e^x )dx =∫_1 ^∞ ((log(1+t))/t^3 ) dt e^x =t =[((log(t+1))/(−2t^2 ))]_1 ^∞ +∫_1 ^∞ (1/(2t^2 (t+1)))dt =((log2)/2)+∫_1 ^∞ (1/(2t^2 ))−(1/(2t))+(1/(2(t+1)))dt =((log2)/2)+(1/2)+[(1/2)log(((t+1)/t))]_1 ^∞ =(1/2)

∫0∞e−2xlog(1+ex)dx=∫1∞log(1+t)t3dtex=t=[log(t+1)−2t2]1∞+∫1∞12t2(t+1)dt=log22+∫1∞12t2−12t+12(t+1)dt=log22+12+[12log(t+1t)]1∞=12

Answered by mathmax by abdo last updated on 28/Nov/20

I =∫_0 ^∞ e^(−2x) ln(1+e^x )dx changement e^x =t give I =∫_1 ^∞ t^(−2) ln(1+t)(dt/t) =∫_1 ^∞ t^(−3) ln(1+t)dt =_(by parts) [−(1/2)t^(−2) ln(1+t)]_1 ^∞ +(1/2)∫_1 ^∞ t^(−2) (1/(1+t))dt =((ln(2))/2) +(1/2)∫_1 ^∞ (dt/(t^2 (t+1))) let decompose F(t)=(1/(t^2 (t+1))) F(t)=(a/t)+(b/t^2 ) +(c/(t+1)) b=1 , c=1 ⇒F(t) =(a/t)+(1/t^2 ) +(1/(t+1)) lim_(t→+∞) tF(t)=0 =a+c ⇒a=−1 ⇒F(t)=−(1/t)+(1/t^2 )+(1/(t+1)) ⇒∫_1 ^∞ F(t)dt =∫_1 ^∞ ((1/(t+1))−(1/t))dt +[−(1/t)]_1 ^∞ =[ln∣((t+1)/t)∣]_1 ^∞ +1 =−ln2 +1 ⇒ I =((ln(2))/2)−((ln(2))/2)+(1/2) ⇒ I =(1/2)

I=∫0∞e−2xln(1+ex)dxchangementex=tgiveI=∫1∞t−2ln(1+t)dtt=∫1∞t−3ln(1+t)dt=byparts[−12t−2ln(1+t)]1∞+12∫1∞t−211+tdt=ln(2)2+12∫1∞dtt2(t+1)letdecomposeF(t)=1t2(t+1)F(t)=at+bt2+ct+1b=1,c=1⇒F(t)=at+1t2+1t+1limt→+∞tF(t)=0=a+c⇒a=−1⇒F(t)=−1t+1t2+1t+1⇒∫1∞F(t)dt=∫1∞(1t+1−1t)dt+[−1t]1∞=[ln∣t+1t∣]1∞+1=−ln2+1⇒I=ln(2)2−ln(2)2+12⇒I=12

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