I need to prove that lim n → ∞ ∫ 0 n 2 n sin ⁡ ( x / n ) e − x 2 d x = 1 / 2

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I need to prove that      lim          n      →      ∞            ∫    0                  n        2              n  sin  ⁡  (  x      /    n  )      e          −              x        2              d  x  =  1      /    2

Arely Davila · Accepted Answer

First recall      lim          y      →      0                  sin      ⁡      y        y    =  1.It follows easily that the integrands converge pointwise to   x      e          −              x        2             on   [  0  ,  +  ∞  )Now recall       |    sin  ⁡  y      |    ≤      |    y      |   for all y.So      |    n  sin  ⁡  (  x      /    n  )      e          −              x        2                  |    ≤  x      e          −              x        2            for all   x  ≥  0The latter is integrable on   [  0  ,  +  ∞  )So Lebesgue domninated convergence theorem applies and yields that the limit is equal to      ∫    0          +      ∞        x      e          −              x        2              d  x  =      1    2  by an easy change of variable.

I need to prove that lim_(n-> infty) int_0^(n^2) n sin(x/n)e^(-x^2)dx=1/2

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