Show that f ( x ) = ∑ k = − ∞ ∞ c k e − i k x = 2 sinh ⁡ ( π ) π ∑ k = 1 ∞ ( − 1 ) k − 1 k 1 + k 2 sin ⁡ ( k x )where c k = 1 4 π ∫ − π π e i k x ( e x − e − x ) d x

Question

Show that  f  (  x  )  =      ∑          k      =      −      ∞        ∞        c    k        e          −      i      k      x        =            2      sinh      ⁡      (      π      )        π        ∑          k      =      1        ∞              (      −      1              )                  k          −          1                    k              1      +              k        2              sin  ⁡  (  k  x  )where             c      k        =          1              4        π                    ∫              −        π            π              e              i        k        x              (          e      x        −          e              −        x              )        d    x

Annabella Velez · Accepted Answer

Consider first      A    k    (  x  )  =      1          i      k      +      1            e          (      i      k      +      1      )      x        −      1          i      k      −      1            e          (      i      k      −      1      )      x            A    k    (  x  )  =            i      k      −      1              (      i      k      +      1      )      (      i      k      −      1      )            e          (      i      k      +      1      )      x        −            i      k      +      1              (      i      k      −      1      )      (      i      k      +      1      )            e          (      i      k      −      1      )      x            A    k    (  x  )  =            1      −      i      k              1      +              k        2                  e          (      i      k      +      1      )      x        +            1      +      i      k              1      +              k        2                  e          (      i      k      −      1      )      x        =            e              i        k        x                    1      +              k        2                        (        (  1  −  i  k  )      e          x        +  (  1  +  i  k  )      e          −      x                  )      Now, replace       e          ±      x        =  cosh  ⁡  (  x  )  ±  sinh  ⁡  (  x  ) to get      A    k    (  x  )  =            2              e                  i          k          x                    (      cosh      ⁡      (      x      )      −      i      k      sinh      ⁡      (      x      )      )              1      +              k        2            Now, using the integration bounds      A    k    (  π  )  −      A    k    (  −  π  )  =            4      i      (      cosh      ⁡      (      π      )      sin      ⁡      (      π      k      )      −      k      sinh      ⁡      (      π      )      cos      ⁡      (      π      k      )      )              1      +              k        2            and, since k is an integer,      A    k    (  π  )  −      A    k    (  −  π  )  =  −            4      i      k      sinh      ⁡      (      π      )      cos      ⁡      (      π      k      )              1      +              k        2              =  −            4      i      k      sinh      ⁡      (      π      )      (      −      1              )        k                    1      +              k        2            and hence      c    k    =  −            i      k      (      −      1              )        k                    π      (      1      +              k        2            )        sinh  ⁡  (  π  )

Show that f ( x ) = <munderover> ∑ <mrow class="MJX-TeXAtom-O

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