Does ∫ 0 1 ∑ n = 0 ∞ x e − n x d x = ∑ n = 0 ∞ ∫ 0 1 x e − n x d x ?

Question

Does      ∫    0    1        ∑          n      =      0        ∞    x      e          −      n      x          d  x  =      ∑          n      =      0        ∞        ∫    0    1    x      e          −      n      x        d  x ?

Sawyer Day · Accepted Answer

You can use Fubini&#039;s theorem, but it seems overkill. Note that for all integer N we have      ∑          n      =      0        N        ∫    0    1    x      e          −      n      x        d  x  =      ∫    0    1        ∑          n      =      0        N    x      e          −      n      x        d  x  ≤      ∫    0    1        ∑          n      =      0              +      ∞        x      e          −      n      x        d  x  ,so      ∑          n      =      0              +      ∞            ∫    0    1    x      e          −      n      x        d  x  ≤      ∫    0    1        ∑          n      =      0              +      ∞        x      e          −      n      x        d  x  .For the reversed inequality, fix   ε  &amp;gt;  0. Since       ∑          n      =      0              +      ∞        x      e          −      n      x       is integrable, we can find a   δ  &amp;gt;  0 such that       ∫    0          δ            ∑          n      =      0              +      ∞        x      e          −      n      x        ≤  ε. And the series       ∑          n      =      0              +      ∞        x      e          −      n      x       is normally convergent on   [  δ  ,  1  ]. So we have                              ∫          0          1                          ∑                      n            =            0                                +            ∞                          x                  e                      −            n            x                          d        x                            =                  ∫          0          δ                          ∑                      n            =            0                                +            ∞                          x                  e                      −            n            x                          d        x        +                  ∫          δ          1                          ∑                      n            =            0                                +            ∞                          x                  e                      −            n            x                          d        x                                          ≤        ε        +                  ∫          δ          1                          ∑                      n            =            0                                +            ∞                          x                  e                      −            n            x                          d        x                                          =        ε        +                  ∑                      n            =            0                                +            ∞                                    ∫          δ          1                x                  e                      −            n            x                          d        x                                          ≤        ε        +                  ∑                      n            =            0                                +            ∞                                    ∫          0          1                x                  e                      −            n            x                          d        x        ,            and since   ε is arbitrary we can conclude the equality.

Does <msubsup> ∫ 0 1 </msubsup> <munderover> ∑<!-- ∑

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