Question Number 100514 by mathmax by abdo last updated on 27/Jun/20

calculatelim_(n→+∞)  ∫_0 ^∞  (1−(x/n))^n ln(1+2x)dx

Answered by mathmax by abdo last updated on 27/Jun/20

A_n =∫_0 ^∞  (1−(x/n))^n ln(2x+1)dx =∫_R (1−(x/n))^n ln(2x+1)χ_([0,+∞[) (x)dx  =∫_R f_n (x)dx with f_n (x) =(1−(x/n))^n  ln(2x+1)  f_n  →^(cs)   f(x) =e^(−x) ln(2x+1)  and ∣f_n ∣ ≤f(x) integrable on [0,+∞[   tbeorem of convegence dominee give lim_(n→+∞) A_n =∫_0 ^∞ e^(−x) ln(2x+1)dx =L  by parts  L =[−e^(−x) ln(2x+1)]_0 ^(+∞)  +∫_0 ^∞ e^(−x)  (2/(2x+1))dx  =2 ∫_0 ^∞  (e^(−x) /(2x+1))dx =_(2x+1=t)    2∫_1 ^(+∞)  (e^(−(((t−1)/2))) /t)×(dt/2) =(√e)∫_1 ^(+∞)  (e^(−(t/2)) /t)dt...be continued...

Answered by maths mind last updated on 28/Jun/20

e^(−x) ≥1−x⇔,x>0  ((e^(−x) −1)/x)≥−1...true by mean valut th  f(t)=e^(−t)  over [0,x]  x>0∃c∈[0,x]  ⇒((e^(−x) −1)/x)=−e^(−c) ≥−1  ⇒e^(−(x/n)) ≥1−(x/n)⇒(1−(x/n))^n ln(1+2x)≤e^(−x) ln(1+2x)  ∫_0 ^(+∞) e^(−x) ln(1+2x)dx<∞  true  ⇒ convergence theorem  ⇒lim_(n→∞) ∫_0 ^∞ (1−(x/n))^n ln(1+2x)dx=∫_0 ^(+∞) lim_(n→∞) (1−(x/n))^n ln(1+2x)dx  =∫_0 ^(+∞) e^(−x) ln(1+2x)dx  [−e^(−x) ln(1+2x)]+∫_0 ^(+∞) (e^(−x) /(1+2x))dx  =∫_0 ^(+∞) (e^(−x) /(1+2x))dx=∫_1 ^(+∞) (e^(−(((u−1)/2))) /(2u))du  =((√e)/2)∫_1 ^(+∞) (e^(−(u/2)) /(2u))du=((√e)/2)∫_(1/2) ^(+∞) (e^(−t) /(2t))  =((√e)/4)∫_(1/2) ^(+∞) (e^(−t) /t)dt=((√e)/4)E((1/2))