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Question Number 35551 by tanmay.chaudhury50@gmail.com last updated on 20/May/18

Commented by abdo mathsup 649 cc last updated on 21/May/18

let find the vslue of ∫_0 ^∞ ((e^(−ax) −e^(−bx) )/x)dx with a>0,b>0 let consider f(t) =∫_0 ^∞ ((e^(−ax) −e^(−bx) )/x)e^(−tx) d3 witht>0 after verifying the condition of derivsbility we have f^′ (t) = −∫_0 ^∞ e^(−tx) { e^(−ax) −e^(−bx) }dx =∫_0 ^∞ {e^(−(b+t)x) −e^(−(a+t)x) }dx =[((−1)/(b+t)) e^(−(b+t)x) + (1/(a+t)) e^(−(a+t)x) ]_0 ^(+∞) = (1/(b+t)) −(1/(a+t)) ⇒ f(t) = ln∣((b+t)/(a+t))∣ +c but ∃m>0 / ∣f(t)∣ ≤ m ∫_0 ^∞ e^(−tx) dx =(m/t) →0(t→+∞) c =lim_(t→+∞) { f(t) −ln∣((b+t)/(a+t))∣} =0⇒ f(t) =ln (((b+t)/(a+t))) and ∫_0 ^∞ ((e^(−ax) −e^(−bx) )/x)dx =f(0) =ln((b/a)) .

letfindthevslueof∫0∞e−ax−e−bxxdxwitha>0,b>0letconsiderf(t)=∫0∞e−ax−e−bxxe−txd3witht>0afterverifyingtheconditionofderivsbilitywehavef′(t)=−∫0∞e−tx{e−ax−e−bx}dx=∫0∞{e−(b+t)x−e−(a+t)x}dx=[−1b+te−(b+t)x+1a+te−(a+t)x]0+∞=1b+t−1a+t⇒f(t)=ln∣b+ta+t∣+cbut∃m>0/∣f(t)∣⩽m∫0∞e−txdx=mt→0(t→+∞)c=limt→+∞{f(t)−ln∣b+ta+t∣}=0⇒f(t)=ln(b+ta+t)and∫0∞e−ax−e−bxxdx=f(0)=ln(ba).

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