Antiderivative of periodic functionI practise before the real analysis exam and I got stuck on few excercises."if F is an antiderivative of a continuous periodic function and F is bounded, then F is periodic" → one should give a proof or counterexample.(I'm convinced it's true, but how to prove it?)

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Antiderivative of periodic functionI practise before the real analysis exam and I got stuck on few excercises.&quot;if F is an antiderivative of a continuous periodic function and F is bounded, then F is periodic&quot;     →    one should give a proof or counterexample.(I&#039;m convinced it&#039;s true, but how to prove it?)

Kendrick Fritz · Accepted Answer

Step 1What is special about   1  +  sin  ⁡  (  x  ) that gives it the following property?      lim          R      →      ∞            ∫    0          R        1  +  sin  ⁡  (  x  )     d  x  =  ∞Step 2If this integral was to have a maximum, what points would you look for first?Further,       d          d      R            ∫    0          R        f  (  x  )  d  x  =  f  (  R  )

Klanglinkmgk · Accepted Answer

Step 1Start with the definition of the antiderivative(s) of f:  F  (  x  )  =      ∫    0    x    f  (  t  )  d  t  +  Cwhere C is a constant. Since f is periodic, with period   T  &amp;gt;  0, then  F  (  2  T  )  =      ∫    0          2      T        f  (  t  )  d  t  +  C  =      ∫    0          T        f  (  t  )  d  t  +      ∫    T          2      T        f  (  t  )  d  t  +  C  =  2      ∫    0          T        f  (  t  )  d  t  +  C  F  (  3  T  )  =      ∫    0          3      T        f  (  t  )  d  t  +  C  =  F  (  2  T  )  +      ∫    T          2      T        f  (  t  )  d  t  =  3      ∫    0          T        f  (  t  )  d  t  +  CStep 2Can you generalize this to find an expression for F(kT) for any integer k (for example by induction)? What does this tell you about       ∫    0          T        f  (  t  )  d  t given the conditions F have to satisfy?Finally what is the value of   F  (  x  +  T  )  −  F  (  x  )  =      ∫    x          x      +      T        f  (  t  )  d  t  =      ∫    x          T        f  (  t  )  d  t  +      ∫    T          x      +      T        f  (  t  )  d  t given the result found above?

Antiderivative of periodic function I practise before the real analysis exam and I got stuck on few

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New Questions in Calculus 1