We are given an integrable function f : [ 0 , 1 ] → R and a set A = { ( x , y ) ∈ R 2 | x , y ≥ 0 , x + y ≤ 1 }. I need to show that for α , β &gt; 0 the following holds: ∫ A f ( x + y ) x α y β d λ 2 ( x , y ) = ∫ 0 1 y α ( 1 − y ) β d λ ( y ) ∫ 0 1 f ( x ) x α + β + 1 d λ ( x ) .We are provided with the hint to use the following map ϕ : [ 0 , 1 ] 2 → R 2 , ϕ ( x , y ) = ( x y , x ( 1 − y ) ) .. I do not see how to use the hint. It is clear that in addition to f all other functions that one finds in the integrals on both sides are integrable on the set A and their product as well. Thus, once one can figure out how to separate the variables, the claim will follow by using the Fubini theorem.Can somebody provide some insight or a solution proposal? Thanks.

Question

We are given an integrable function   f  :  [  0  ,  1  ]  →      R   and a set   A  =  {  (  x  ,  y  )  ∈            R        2        |    x  ,  y  ≥  0  ,  x  +  y  ≤  1  }. I need to show that for   α  ,  β  &amp;gt;  0 the following holds:      ∫    A    f  (  x  +  y  )      x          α            y          β        d      λ    2    (  x  ,  y  )  =      ∫    0    1        y          α        (  1  −  y      )          β        d  λ  (  y  )      ∫    0    1    f  (  x  )      x          α      +      β      +      1        d  λ  (  x  )  .We are provided with the hint to use the following map   ϕ  :  [  0  ,  1      ]    2    →            R        2    ,  ϕ  (  x  ,  y  )  =  (  x  y  ,  x  (  1  −  y  )  )  .. I do not see how to use the hint. It is clear that in addition to   f all other functions that one finds in the integrals on both sides are integrable on the set   A and their product as well. Thus, once one can figure out how to separate the variables, the claim will follow by using the Fubini theorem.Can somebody provide some insight or a solution proposal? Thanks.

Nirdaciw3 · Accepted Answer

The transformation   (  x  ,  y  )  ↦      (    x    +    y    ,          x              x        +        y              )    =  (  u  ,  v  ) maps   A to   (  0  ,  1      )    2   bijectively and has inverse   (  u  ,  v  )  ↦  (  u  v  ,  u  (  1  −  v  )  )  =  (  x  ,  y  ). Moreover,            ∂      (      u      ,      v      )              ∂      (      x      ,      y      )        =  −      1          x      +      y        =  −      1    u  Therefore,                              ∬                      A                          f        (        x        +        y        )                  x                      α                                    y                      β                                    d                A                            =                              ∬                      (            0            ,            1                          )              2                                      f        (        u        )        (        u        v                  )                      α                                                [            u            (            1            −            v            )            ]                                β                          u                  d                A                                          =                              ∬                      (            0            ,            1                          )              2                                      f        (        u        )                  u                      α            +            β            +            1                                    v                      α                          (        1        −        v                  )                      β                                    d                A                                          =                              ∫          0          1                f        (        x        )                  x                      α            +            β            +            1                                    d                x        ⋅                  ∫          0          1                          x                      α                          (        1        −        x                  )                      β                                    d                x

We are given an integrable function f : [ 0 , 1 ] → <mr

Answered question

Answer & Explanation

New Questions in Pre-Algebra