So I have to evaluate ∫ − ∞ ∞ e a x cosh ⁡ x d xI tried takeing the analytic expansion, and integrating over the real axis. I took this as being a half circle from − ∞ to ∞, minus the arc of the circle. over the arc I proved that the integral is zero, and I have left only with a sum of the residues of the function. With the help of a previous question I calculated the values of the residues, but I could not manage to converge the sum.

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So I have to evaluate      ∫          −      ∞        ∞              e              a        x                    cosh      ⁡      x          d  xI tried takeing the analytic expansion, and integrating over the real axis. I took this as being a half circle from   −  ∞ to   ∞, minus the arc of the circle. over the arc I proved that the integral is zero, and I have left only with a sum of the residues of the function. With the help of a previous question I calculated the values of the residues, but I could not manage to converge the sum.

lipovicai1w · Accepted Answer

The usual trick for this integral is to take a rectangular contour with vertices   ±  R and   ±  R  +  π  i and let   R  →  ∞. This contains only one pole, simple at   z  =  π  i      /    2, and the integral over the top edge is closely related to that over the bottom edge (the one you care about).

Deon Moran · Accepted Answer

Assuming   a  ∈  (  −  1  ,  1  ) such integral equals                    (1)                              ∫                      0                                +            ∞                                                              e                              a                x                                      +                          e                              −                a                x                                                          cosh            ⁡            x                                  d        x        =        2                  ∫                      0                                +            ∞                                                cosh            ⁡            (            a            x            )                                cosh            ⁡            x                                  d        x        =                  π                      cos            ⁡                                          π                a                            2                                          for instance by exploiting                    (2)                              ∫                      0                                +            ∞                          cosh        ⁡        (        a        x        )                  e                      −            (            2            m            +            1            )            x                                  d        x        =                              (            2            m            +            1            )                                (            2            m            +            1                          )              2                        −                          a              2                                                              (3)                              ∑                      m            ≥            0                                                (            2            m            +            1            )            (            −            1                          )              m                                            (            2            m            +            1                          )              2                        −                          a              2                                                                          =                                      Herglotz trick                                                π                      4            cos            ⁡                                          π                a                            2                                      .

how to calculate int^infty_(−infty) (e^(ax))/(cos h x)dx

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