The antiderivative of sin(1/x)How to prove that the function f ( x ) = sin ⁡ 1 x for x ≠ 0 , f ( 0 ) = 0 has an antiderivative? This means F ( x ) = ∫ 0 x sin ⁡ ( 1 / t ) d t has derivative 0 at x = 0, but I have no idea how to prove it.

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The antiderivative of sin(1/x)How to prove that the function   f  (  x  )  =  sin  ⁡      1    x   for   x  ≠  0  ,  f  (  0  )  =  0 has an antiderivative? This means   F  (  x  )  =      ∫          0              x        sin  ⁡  (  1      /    t  )  d  t has derivative 0 at   x  =  0, but I have no idea how to prove it.

hadnya1qd · Accepted Answer

Step 1I just blogged another solution to this question in my blog. The idea is to consider the following function  G  (  x  )  =      {                                        x            2                    cos          ⁡                      1            x                    ,                           if           x          ≠          0          ,                                      0          ,                           if           x          =          0.                        which is differentiable. Clearly,      G    ′    (  x  )  =      {                            sin          ⁡                      1            x                    +          2          x          cos          ⁡                      1            x                    ,                           if           x          ≠          0          ,                                      0          ,                           if           x          =          0.                        Step 2Hence,       G    ′    =  f  +  h where  h  (  x  )  =      {                            2          x          cos          ⁡                      1            x                    ,                           if           x          ≠          0          ,                                      0          ,                           if           x          =          0.                        Since h is continuous, it has antiderivative H, thus giving us   f  =  (  G  −  H      )    ′  . In other words,   G  −  H is an antiderivative of f.

wellnesshaus4n4 · Accepted Answer

Explanation:We can substitute   u  =      t          −      1       to get a more convenient expression:                              |                                    F              (              x              )              −              F              (              0              )                        x                    |                                    =                  1                      |            x            |                                    |                      ∫            0            x                    sin          ⁡          (                      t                          −              1                                )                    d          t          |                                    (symmetry)                                  =                  1                      |            x            |                                    |                      ∫                          1                              /                            |              x              |                        ∞                                              sin              ⁡              u                                      u              2                                          d          u          |                                                  =                  1                      |            x            |                                    |                                    [              −                                                cos                  ⁡                  u                                                  u                  2                                            ]                                      1                              /                            |              x              |                        ∞                    −          2                      ∫                          1                              /                            |              x              |                        ∞                                              cos              ⁡              u                                      u              3                                          d          u          |                                                  =                  1                      |            x            |                                    |          |          x                      |            2                    cos          ⁡                      1                          |              x              |                                −          2                      ∫                          1                              /                            |              x              |                        ∞                                              cos              ⁡              u                                      u              3                                          d          u          |                                                  ⩽        |        x        |        +                  1                      |            x            |                                    ∫                      1                          /                        |            x            |                    ∞                          2                      u            3                                  d        u                                          =        2        |        x        |        .            Thus       F    ′    (  0  )  =  0

The antiderivative of sin(1/x) How to prove that the function f <mo stretchy="false">( x <

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