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Question Number 53270 by Abdo msup. last updated on 19/Jan/19

1)calculate ∫_0 ^∞ e^(−at) dt with a>0 2)by using fubinni theorem find the value of ∫_0 ^∞ ((e^(−t) −e^(−xt) )/t)dt with x>0 .

1)calculate∫0∞e−atdtwitha>0 2)byusingfubinnitheoremfindthevalueof ∫0∞e−t−e−xttdtwithx>0.

Commented bymaxmathsup by imad last updated on 20/Jan/19

1) ∫_0 ^∞ e^(−at) dt =[−(1/a) e^(−at) ]_0 ^(+∞) =(1/a) 2) ⇒∫_1 ^x (da/a) =ln(x) we take x>0 but ∫_1 ^x (da/a) =∫_1 ^x (∫_0 ^∞ e^(−at) dt)da =∫_0 ^∞ (∫_1 ^x e^(−at) da)dt ( fubini theorem) =∫_0 ^∞ ( [−(1/t)e^(−at) ]_(a=1) ^(a=x) )dt =∫_0 ^∞ ((e^(−t) −e^(−xt) )/t) dt ⇒ ∫_0 ^∞ ((e^(−t) −e^(−xt) )/t) dt =ln(x) with x>0

1)∫0∞e−atdt=[−1ae−at]0+∞=1a 2)⇒∫1xdaa=ln(x)wetakex>0but ∫1xdaa=∫1x(∫0∞e−atdt)da=∫0∞(∫1xe−atda)dt(fubinitheorem) =∫0∞([−1te−at]a=1a=x)dt=∫0∞e−t−e−xttdt⇒ ∫0∞e−t−e−xttdt=ln(x)withx>0

Answered by kaivan.ahmadi last updated on 19/Jan/19

1) u=−at⇒du=−adt ((−1)/a)∫e^u du=((−1)/a)e^u =((−1)/a)e^(−at) ∣_0 ^∞ =((−1)/a)(e^(−∞) −e^0 )= ((−1)/a)(0−1)=(1/a)

1)u=−at⇒du=−adt −1a∫eudu=−1aeu=−1ae−at∣0∞=−1a(e−∞−e0)= −1a(0−1)=1a

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