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Question Number 182219 by SANOGO last updated on 05/Dec/22

Answered by Ar Brandon last updated on 06/Dec/22

Sea I(α)=∫_0 ^1 ((t^α −1)/(lnt))dt Derivando con respecto a α obtenemos I ′(α)=∫_0 ^1 ((t^α lnt)/(lnt))dt=∫_0 ^1 t^α dt=[(t^(α+1) /(α+1))]_0 ^1 =(1/(α+1)) Ahora integrando con respecto a α obtenemos I(α)=∫(1/(α+1))dα=ln(α+1)+C Pero I(0)=∫_0 ^1 ((t^0 −1)/(lnt))dt=0=ln(0+1)+C ⇒C=0 ⇒I(α)=ln(α+1) ∫_0 ^1 ((t−1)/(lnt))dt=I(1)=ln(2)

SeaI(α)=∫01tα−1lntdtDerivandoconrespectoaαobtenemosI′(α)=∫01tαlntlntdt=∫01tαdt=[tα+1α+1]01=1α+1AhoraintegrandoconrespectoaαobtenemosI(α)=∫1α+1dα=ln(α+1)+CPeroI(0)=∫01t0−1lntdt=0=ln(0+1)+C⇒C=0⇒I(α)=ln(α+1)∫01t−1lntdt=I(1)=ln(2)

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