Define I ( x ) = ∫ x ∞ e − y 2 d yEvaluate ∫ 0 ∞ e x 2 [ I ( x ) ] 2 d x

Question

Define  I  (  x  )  =      ∫    x    ∞        e          −              y        2                 d  yEvaluate      ∫    0    ∞        e                  x        2              [  I  (  x  )      ]    2       d  x

Nadia Berry · Accepted Answer

To wit  I  (  x  )  =  x      ∫    1          ∞        d  u        e          −              x        2                    u        2            which means that  I  (  x      )    2    =      x    2        ∫    1          ∞        d  u        ∫    1          ∞        d  v        e          −              x        2            (              u        2            +              v        2            )      Now we can put this into the desired integral and reverse the order of integration to get                              ∫          0                      ∞                          d        x                          e                                    x              2                                      I        (        x                  )          2                =                  ∫          1                      ∞                          d        u                          ∫          1                      ∞                          d        v                          ∫          0                      ∞                          d        x                          x          2                          e                      −                          x              2                        (                          u              2                        +                          v              2                        −            1            )                              The inner integral converges because       u    2    +      v    2    −  1  ≥  0. We evaluate the inner integral and reduce the integral to a double integral:      ∫    0          ∞        d  x        e                  x        2              I  (  x      )    2    =            π        4        ∫    1          ∞        d  u        ∫    1          ∞        d  v    (      u    2    +      v    2    −  1      )          −      3              /            2      The integral over v may be attacked by a trig substitution:   v  =            u      2        −    1    tan  ⁡      t  ,   d  v  =            u      2        −    1        sec    2    ⁡      t    d  t; the result is      ∫    1          ∞        d  v    (      u    2    +      v    2    −  1      )          −      3              /            2        =      (    1    −          1      u        )        1                  u        2            −      1        =      1          u      (      u      +      1      )      The problem now reduces to the evaluation of      ∫    1          ∞                  d      u              u      (      u      +      1      )        =      lim          r      →      ∞        log  ⁡            (              r                  r          +          1                    )        +  log  ⁡      2    =  log  ⁡      2  Therefore, the desired integral has the value      ∫    0          ∞        d  x        e                  x        2              I  (  x      )    2    =            π        4    log  ⁡      2

Define I(x)=int_x^infty e^(-y^2)dy. Evaluate int_0^infty e^(x^2)[I(x)]^2dx

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