Evaluate ∫ 0 ∞ sin 2 ⁡ ( t ) t 2 d t with help of Laplace transformUsing the following identity ∫ 0 ∞ f ( t ) t d t = ∫ 0 ∞ L { f ( t ) } ( u ) d uI rewrote ∫ 0 ∞ sin 2 ⁡ ( t ) t 2 d tas ∫ 0 ∞ sin 2 ⁡ ( t ) t ⋅ t d tAnd thus the initial integral should be easily evaluated as ∫ 0 ∞ L { sin 2 ⁡ t t } ( u ) d uAccording to my calculations, this is equal to ∫ 0 ∞ ( 1 4 ln ⁡ ( u 2 + 4 ) − u 2 4 ) d uWhich evaluates to π 16 . Being that this acutally a well-known integral and that its value is actually π 2 I think that I made a crucial mistake somewhere. Any ideas?

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Evaluate       ∫          0              ∞                          sin                  2                    ⁡              (        t        )                    t              2              d  t with help of Laplace transformUsing the following identity      ∫          0              ∞                  f              (        t        )              t    d  t  =      ∫          0              ∞            L        {    f          (      t      )        }        (    u    )    d  uI rewrote      ∫          0              ∞                          sin                  2                    ⁡              (        t        )                    t              2              d  tas      ∫          0              ∞                          sin                  2                    ⁡              (        t        )                    t      ⋅      t        d  tAnd thus the initial integral should be easily evaluated as      ∫          0              ∞            L        {                            sin                      2                          ⁡        t            t        }        (    u    )    d  uAccording to my calculations, this is equal to      ∫          0              ∞            (          1      4        ln    ⁡          (              u                  2                    +      4      )        −                  u                  2                    4        )    d  uWhich evaluates to       π    16  . Being that this acutally a well-known integral and that its value is actually       π    2   I think that I made a crucial mistake somewhere. Any ideas?

Addison Mueller · Accepted Answer

You seemed to have confused logarithmic identities. The integrand should be      1    4    log  ⁡      (          4              s        2              +    1    )  which indeed evaluates as   π      /    2. I am guessing you wrongly factored out an       s    2

siotaody · Accepted Answer

We have                              ∫          0                      ∞                                                              sin              2                        ⁡            t                    t                          e                      −            u            t                                              =                  1          2                          ∫          0                      ∞                                                1            −            cos            ⁡            t                    t                          e                      −            (            u                          /                        2            )            t                          d        t                                                        =                  1          2                log        ⁡                  (                                    (              u                              /                            2                              )                2                            +              1                                      (              u                              /                            2                              )                2                                              )                                                                =                  1          2                log        ⁡                  (                                                    u                2                            +              4                                      u              2                                )

Evaluate int_0^(oo) (sin^2(t))/(t^2) dt with help of Laplace transform

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