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Question Number 107247 by ZiYangLee last updated on 09/Aug/20

Answered by abdomathmax last updated on 09/Aug/20

let f(t) =∫_0 ^∞ ((e^(−x) −e^(−2x) )/x) e^(−tx) dx with t≥0 we hsve f^′ (t) =−∫_0 ^∞ (e^(−x) −e^(−2x) )e^(−tx) dx =∫_0 ^∞ ( e^(−(t+2)x) −e^(−(t+1)x) )dx =[−(1/(t+2))e^(−(t+2)x) +(1/(t+1))e^((t+1)x) ]_0 ^∞ =(1/(t+2))−(1/(t+1)) ⇒f(t) =ln(((t+2)/(t+1))) +C ∃m>0 / ∣f(t)∣≤(m/t) →0 (t→+∞) ⇒C =0 ⇒ f(t) =ln(((t+2)/(t+1))) t=0 weget ∫_0 ^∞ ((e^(−x) −e^(−2x) )/x)dx =ln(2)

letf(t)=∫0∞e−x−e−2xxe−txdxwitht⩾0wehsvef′(t)=−∫0∞(e−x−e−2x)e−txdx=∫0∞(e−(t+2)x−e−(t+1)x)dx=[−1t+2e−(t+2)x+1t+1e(t+1)x]0∞=1t+2−1t+1⇒f(t)=ln(t+2t+1)+C∃m>0/∣f(t)∣⩽mt→0(t→+∞)⇒C=0⇒f(t)=ln(t+2t+1)t=0weget∫0∞e−x−e−2xxdx=ln(2)

Commented by ZiYangLee last updated on 09/Aug/20

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