Verifying Fourier series of antiderivative of a functionThe function is f ( x ) = { 0 , − π ≤ x ≤ 0 x , 0 &lt; x ≤ π .The Fourier series of the function is: 1 2 ( π 2 ) + ∑ n = 1 ∞ ( − 1 ) n − 1 π n 2 cos ⁡ ( n x ) + ( − 1 ) n + 1 n sin ⁡ ( n x ).Since the function can be integrated termwise, the Fourier series for the antiderivative ∫ − π x f ( x ) d x is: 1 2 ( π 2 ) ( x + π ) + ∑ n = 1 ∞ 1 n ( ( − 1 ) n − 1 π n 2 sin ⁡ ( n x ) + ( − 1 ) n n ( cos ⁡ ( n x ) − ( − 1 ) n ) ) The antiderivative of the function is F ( x ) = { 0 , − π ≤ x ≤ 0 x 2 2 , 0 &lt; x ≤ π . The Fourier series of the antiderivative is: 1 2 ( π 2 6 ) + ∑ n = 1 ∞ 1 n ( ( ( − 1 ) n − 1 π n 2 − ( − 1 ) n π 2 ) sin ⁡ ( n x ) + ( − 1 ) n n cos ⁡ ( n x ) ) Where I am missing? Why I am having wrong answer. Even if I did some mistakes in calculations, what's with x term obtained in piece wise integration. Thanks in advance.

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Verifying Fourier series of antiderivative of a functionThe function is   f  (  x  )  =      {                            0          ,                          −          π          ≤          x          ≤          0                                      x          ,                          0          &amp;lt;          x          ≤          π                        .The Fourier series of the function is:      1    2    (      π    2    )  +      ∑          n      =      1        ∞              (      −      1              )        n            −      1              π              n                  2                      cos  ⁡  (  n  x  )  +            (      −      1              )                  n          +          1                      n    sin  ⁡  (  n  x  ).Since the function can be integrated termwise, the Fourier series for the antiderivative       ∫          −      π        x    f  (  x  )    d  x is:      1    2    (      π    2    )  (  x  +  π  )  +      ∑          n      =      1        ∞        1    n        (                  (        −        1                  )          n                −        1                    π                  n                      2                                sin    ⁡    (    n    x    )    +                  (        −        1                  )                      n                              n        (    cos    ⁡    (    n    x    )    −    (    −    1          )      n        )    )  The antiderivative of the function is   F  (  x  )  =      {                            0          ,                          −          π          ≤          x          ≤          0                                                                x              2                        2                    ,                          0          &amp;lt;          x          ≤          π                        . The Fourier series of the antiderivative is:      1    2    (            π      2        6    )  +      ∑          n      =      1        ∞        1    n        (          (                        (          −          1                      )            n                    −          1                          π                      n                          2                                          −      (      −      1              )        n                    π        2            )        sin    ⁡    (    n    x    )    +                  (        −        1                  )                      n                              n        cos    ⁡    (    n    x    )    )  Where I am missing? Why I am having wrong answer. Even if I did some mistakes in calculations, what&#039;s with x term obtained in piece wise integration. Thanks in advance.

Rose Holmes · Accepted Answer

Step 1From comments, What you have is:      π    4    x  +            π      2        4    +  ∑      1    n        (                  (        −        1                  )          n                −        1                    π                  n          2                      sin    ⁡    (    n    x    )    +                  (        −        1                  )          n                    n        cos    ⁡    (    n    x    )    )    −  ∑      1          n      2      The Fourier expansion of x on   [  −  π  ,  π  ] is   x  =  −  2  ∑            (      −      1              )        n              n    sin  ⁡  (  n  x  ) and using the fact that   ∑      1          n      2        =            π      2        6  .Step 2It is obttained that,       π    4        (    −    2    ∑                  (        −        1                  )          n                    n        sin    ⁡    (    n    x    )    )    +            π      2        4    +  ∑      1    n        (                  (        −        1                  )          n                −        1                    π                  n          2                      sin    ⁡    (    n    x    )    +                  (        −        1                  )          n                    n        cos    ⁡    (    n    x    )    )    −            π      2        6              π      2        12    +  ∑      1    n        (          (                        (          −          1                      )            n                    −          1                          π                      n            2                              −      (      −      1              )        n                    π        2            )        sin    ⁡    (    n    x    )    +                  (        −        1                  )          n                    n        cos    ⁡    (    n    x    )    )  Now both results conicides.

Verifying Fourier series of antiderivative of a function

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