Antiderivative of sin ⁡ ( 1 x )I have to find the values of parameter a so the function f : R → R , f ( x ) = { sin ⁡ 1 x for x ∈ R ∖ { 0 } a for x = 0 . has an antiderivative on R . I know that lim x → 0 sin ⁡ 1 x does not exist in x = 0. I have found that the answer is a = 0 but i can't explain it.

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Antiderivative of   sin  ⁡  (      1    x    )I have to find the values of parameter a so the function  f  :      R    →      R    ,  f  (  x  )  =      {                            sin          ⁡                      1            x                                     for           x          ∈                      R                    ∖          {          0          }                                      a                           for           x          =          0          .                        has an antiderivative on       R  . I know that       lim          x      →      0        sin  ⁡      1    x   does not exist in   x  =  0. I have found that the answer is   a  =  0 but i can&#039;t explain it.

Jerome Page · Accepted Answer

Step 1Suppose F is a primitive of f. Since f is continuous on       R    ∖  {  0  }, we have  F  (  y  )  −  F  (  x  )  =      ∫    x    y    f  (  t  )    d  tfor   0  &amp;lt;  x  &amp;lt;  y or   x  &amp;lt;  y  &amp;lt;  0 by the fundamental theorem of calculus. The continuity of F then forces  F  (  x  )  =  F  (  0  )  +      ∫    0    x    sin  ⁡      1    t      d  tfor all   x  ∈      R  . Thus it remains to see that F is differentiable at 0 with derivative 0. Now we have      d          d      t                  (            t    2    cos  ⁡  (      t          −      1        )            )        =  sin  ⁡  (      t          −      1        )  +  2  t  cos  ⁡  (      t          −      1        )and so       ∫    0    x    sin  ⁡  (      t          −      1        )    d  t  =      ∫    0    x        d          d      t                  (            t    2    cos  ⁡  (      t          −      1        )            )        −  2  t  cos  ⁡  (      t          −      1        )    d  t  =      x    2    cos  ⁡  (      x          −      1        )  −      ∫    0    x    2  t  cos  ⁡  (      t          −      1        )    d  t  .Step 2The first part       x    2    cos  ⁡  (      x          −      1        ) is easily seen to be differentiable at 0 with derivative 0 there, and since  g  (  t  )  =      {                            0                          ,          t          =          0                                      2          t          cos          ⁡          (                      t                          −              1                                )                          ,          t          ≠          0                        is continuous on all of       R  , the fundamental theorem of calculus asserts its differentiability on       R  , and the derivative is g. Hence       F    ′    (  0  )  =  0  −  g  (  0  )  =  0.

Antiderivative of sin ⁡ ( 1 x </mfrac> ) I have to

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