I would like to find an equivalent of the sequence u n where u n = ∫ 0 π / 2 cos ⁡ ( π 2 sin ⁡ ( x ) ) n d x

Question

I would like to find an equivalent of the sequence       u          n       where      u          n        =      ∫    0          π              /            2        cos  ⁡            (              π        2            sin      ⁡      (      x      )      )        n        d    x

Crevani9a · Accepted Answer

For you to conclude             u      n        ∼                  2                  n          π                     you will need to show  I  (  n  )  =      ∫    0          π              /            2            cos    n    ⁡  (  x  )      (          1              1        −        4                  x          2                          /                          π          2                      −    1    )    d  x  =      o        (          1              n              )  i.e not just       o    (  1  )  . The main contribution to the integral is near   x  =  0 and the integrand drops off exponentially away from there. Near   x  =  0 we have (by Taylor expansion)       cos    n    ⁡  (  x  )  ≈  (  1  −      x    2        /    2      )    n    ≈  exp  ⁡  (  −  n      x    2        /    2  ) and             1              1        −        4                  x          2                          /                          π          2                      −    1    ≈    2          x      2              /              π      2        I  (  n  )  ≈      ∫    0          ∞                  2              x        2                    π      2        exp  ⁡  (  −  n      x    2        /    2  )  d  x  =            2                        n          3                          π          3                      .Thus one would reasonably conjecture that      u    n    =            2              n        π                  (    1    +          2              n                  π          2                      +          O              (              1                  n          2                    )        )    .

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