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find-by-two-ways-the-value-of-0-1-x-y-dxdxy-then-calculate-0-1-t-1-lnt-dt-

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Question Number 27692 by abdo imad last updated on 12/Jan/18
find by two ways the value of ∫∫_([0,1]) x^y dxdxy then calculate ∫_0 ^1 ((t−1)/(lnt))dt .
findbytwowaysthevalueof∫∫[0,1]xydxdxythencalculate∫01t−1lntdt.
Commented by abdo imad last updated on 13/Jan/18
let put I= ∫∫_([0,1]^2 ) x^y dxdy I= ∫_0 ^1 (∫_0 ^1 ( e^(ylnx) dy)dx but ∫_0 ^1 e^(ylnx) dy = [(1/(lnx)) x^y ]_(y=0) ^(y=1) = ((x−1)/(lnx)) ⇒ I= ∫_0 ^1 ((x−1)/(lnx))dx let change the integral due to fubini theorem I=∫_0 ^1 ( ∫_0 ^1 x^y dx)dy but ∫_0 ^1 x^y dx =[ (x^(y+1) /(y+1))]_(x=0) ^(x=1) = (1/(y+1)) ⇒I= ∫_0 ^1 (dy/(y+1)) =[ln/y+1/]_0 ^1 =ln(2).finally ∫_0 ^1 ((x−1)/(lnx))dx= ln(2).
letputI=∫∫[0,1]2xydxdyI=∫01(∫01(eylnxdy)dxbut∫01eylnxdy=[1lnxxy]y=0y=1=x−1lnx⇒I=∫01x−1lnxdxletchangetheintegralduetofubinitheoremMissing \left or extra \right=1y+1⇒I=∫01dyy+1=[ln/y+1/]01=ln(2).finally∫01x−1lnxdx=ln(2).

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