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Calculate-when-a-b-are-positive-reals-f-a-b-0-1-t-a-t-b-lnt-dt-




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)=01tatblntdt
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+eaxebxx(ex)dx=0e(a+1)xe(b+1)xxdx=0e(b+1)xe(a+1)xxdxwehavefa(a,b)=0xe(a+1)xxdx=0e(a+1)xdx=[1a+1e(a+1)x]0+=1a+1{1}=1a+1f(a,b)=ln(a+1)+cwehavef(0,b)=c=011tblntdt=lnt=u01ebtu(eu)du=0eue(b+1)uudu=0e(b+1)ueuudu=φ(b)wehaveφ(b)=0e(b+1)udu=[1b+1e(b+1)u]0+=1b+1φ(b)=ln(b+1)+cwithc=φ(0)=0f(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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