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Question Number 70909 by rajesh4661kumar@gmail.com last updated on 09/Oct/19

Commented by rajesh4661kumar@gmail.com last updated on 10/Oct/19

28

28

Commented by Abdo msup. last updated on 10/Oct/19

f(t)=∫_0 ^1 ((x^t −1)/(lnx))dx changement lnx=−u give f(t)=−∫_0 ^∞ ((e^(−tu) −1)/(−u))(−e^(−u) )du =−∫_0 ^∞ ((e^(−(t+1)u) −e^(−u) )/u) du=∫_0 ^∞ ((e^(−u) −e^(−(t+1)u) )/u)du f^′ (t) = ∫_0 ^∞ e^(−(t+1)u) du =[−(1/(t+1))e^(−(t+1)u) ]_(u=0) ^∞ =(1/(t+1)) ⇒f(t)=ln(t+1) +c f(o)=0=c ⇒f(t)=ln(t+1) (t≥0)

f(t)=∫01xt−1lnxdxchangementlnx=−ugivef(t)=−∫0∞e−tu−1−u(−e−u)du=−∫0∞e−(t+1)u−e−uudu=∫0∞e−u−e−(t+1)uuduf′(t)=∫0∞e−(t+1)udu=[−1t+1e−(t+1)u]u=0∞=1t+1⇒f(t)=ln(t+1)+cf(o)=0=c⇒f(t)=ln(t+1)(t⩾0)

Answered by Henri Boucatchou last updated on 10/Oct/19

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