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Question Number 67462 by ~ À ® @ 237 ~ last updated on 27/Aug/19

Calculate when a,b are positive reals f(a,b)= ∫_0 ^1 ((t^a −t^b )/(lnt)) dt

Calculatewhena,barepositiverealsf(a,b)=∫01ta−tblntdt

Commented by mathmax by abdo last updated on 27/Aug/19

changement lnt =−x give f(a,b) =−∫_0 ^(+∞) ((e^(−ax) −e^(−bx) )/(−x))(−e^(−x) )dx =−∫_0 ^∞ ((e^(−(a+1)x) −e^(−(b+1)x) )/x)dx =∫_0 ^∞ ((e^(−(b+1)x) −e^(−(a+1)x) )/x)dx we have (∂f/∂a)(a,b) =∫_0 ^∞ ((xe^(−(a+1)x) )/x)dx =∫_0 ^∞ e^(−(a+1)x) dx =[((−1)/(a+1)) e^(−(a+1)x) ]_0 ^(+∞) =−(1/(a+1)){−1} =(1/(a+1)) ⇒ f(a,b) =ln(a+1) +c we have f(0,b) =c =∫_0 ^1 ((1−t^b )/(lnt))dt =_(lnt =−u) −∫_0 ^∞ ((1−e^(−bt) )/(−u))(−e^(−u) )du =−∫_0 ^∞ ((e^(−u) −e^(−(b+1)u) )/u)du =∫_0 ^∞ ((e^(−(b+1)u) −e^(−u) )/u)du =ϕ(b) we have ϕ^′ (b) =−∫_0 ^∞ e^(−(b+1)u) du =[(1/(b+1)) e^(−(b+1)u) ]_0 ^(+∞) =−(1/(b+1)) ⇒ ϕ(b) =−ln(b+1) +c with c=ϕ(0) =0 ⇒ f(0,b) =−ln(b+1) ⇒f(a,b) =ln(a+1)−ln(b+1) =ln(((a+1)/(b+1))).

changementlnt=−xgivef(a,b)=−∫0+∞e−ax−e−bx−x(−e−x)dx=−∫0∞e−(a+1)x−e−(b+1)xxdx=∫0∞e−(b+1)x−e−(a+1)xxdxwehave∂f∂a(a,b)=∫0∞xe−(a+1)xxdx=∫0∞e−(a+1)xdx=[−1a+1e−(a+1)x]0+∞=−1a+1{−1}=1a+1⇒f(a,b)=ln(a+1)+cwehavef(0,b)=c=∫011−tblntdt=lnt=−u−∫0∞1−e−bt−u(−e−u)du=−∫0∞e−u−e−(b+1)uudu=∫0∞e−(b+1)u−e−uudu=φ(b)wehaveφ′(b)=−∫0∞e−(b+1)udu=[1b+1e−(b+1)u]0+∞=−1b+1⇒φ(b)=−ln(b+1)+cwithc=φ(0)=0⇒f(0,b)=−ln(b+1)⇒f(a,b)=ln(a+1)−ln(b+1)=ln(a+1b+1).

Commented by ~ À ® @ 237 ~ last updated on 28/Aug/19

Thanks you sir

Thanksyousir

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