Implication of complex-differentiability on the existence of antiderivativesThis is something that came to me randomly:We know that a complex-differentiable function is infinitely-differentiable, and that elementary complex functions follow the same formal rules for differentiation as elementary real functions.Does this mean that whenever we have a function defined via elementary functions on R whose derivative is not differentiable everywhere on R, we can conclude that this function does not have an antiderivative that can be expressed in terms of elementary functions?For example: the function x 3 s i n ( 1 x 2 ) extended by zero at the origin. This function is differentiable everywhere on R, but its derivative is not continuous (thus not differentiable) at the origin.Thus if x 3 s i n ( 1 x 2 ) had an antiderivative expressible in terms of elementary functions, we would obtain a contradiction with the fact that differentiable complex functions are smooth.Here are my questions:1) Is my reasoning sound?2) Is this discussed somewhere in more generality, and has it been formalized in some way?3) Does anyone have any examples where this observation leads to interesting conclusions?

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Implication of complex-differentiability on the existence of antiderivativesThis is something that came to me randomly:We know that a complex-differentiable function is infinitely-differentiable, and that elementary complex functions follow the same formal rules for differentiation as elementary real functions.Does this mean that whenever we have a function defined via elementary functions on R whose derivative is not differentiable everywhere on R, we can conclude that this function does not have an antiderivative that can be expressed in terms of elementary functions?For example: the function       x    3    s  i  n  (      1          x      2        ) extended by zero at the origin. This function is differentiable everywhere on R, but its derivative is not continuous (thus not differentiable) at the origin.Thus if       x    3    s  i  n  (      1          x      2        ) had an antiderivative expressible in terms of elementary functions, we would obtain a contradiction with the fact that differentiable complex functions are smooth.Here are my questions:1) Is my reasoning sound?2) Is this discussed somewhere in more generality, and has it been formalized in some way?3) Does anyone have any examples where this observation leads to interesting conclusions?

kuncwadi17 · Accepted Answer

Step 1I would say that your reasoning is not sound, or at least that is lacking something.Looking at your example, even though   f  (  x  )  =      {                                        x            3                    sin          ⁡                      1                          x              2                                ,                          x          ≠          0                                      0          ,                          x          =          0                        is differentiable at   x  =  0 as a function       R    →      R  , the corresponding complex function  f  (  z  )  =      {                                        z            3                    sin          ⁡                      1                          z              2                                ,                          z          ≠          0                                      0          ,                          z          =          0                        has an essential singularity at   z  =  0, so it cannot have (complex, holomorphic) anti-derivative on       C  . It does have a holomorphic anti-derivative on       C    ∖  {  0  }, but that&#039;s more of a coincidence. The residue at   z  =  0 happens to be 0. If you take       x    5    sin  ⁡      1          x      2       instead, the corresponding complex function doesn&#039;t even have an anti-derivative on the punctured plane.Step 2Furthermore, starting with   g  (  x  )  =  4      x    3    sin  ⁡      1    x    −      x    2    cos  ⁡      1    x   the same &quot;argument&quot; would show that g doesn&#039;t have an elementary anti-derivative, since        g    ′    (  x  )  =      {                            12                      x            2                    sin          ⁡                      1            x                    −          6          x          cos          ⁡                      1            x                    −          sin          ⁡                      1            x                    ,                          x          ≠          0                                      0          ,                          x          =          0                         is discontinuous at   x  =  0. But this is false, since g is the derivative of       x    4    sin  ⁡      1    x  .

Implication of complex-differentiability on the existence of antiderivatives This is something that

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