Calculate ∫ 0 π cos ⁡ ( k x ) 1 − 2 τ cos ⁡ ( x ) d xfor k ≥ 0 and | τ | &lt; 1 2 . For k=0, I use the reparameterization t = tan ⁡ ( x / 2 ), but I have no idea how to do it for k ≥ 1.

Question

Calculate       ∫    0          π                  cos      ⁡      (      k      x      )              1      −      2      τ      cos      ⁡      (      x      )        d  xfor   k  ≥  0 and       |    τ      |    &amp;lt;      1    2  . For k=0, I use the reparameterization   t  =  tan  ⁡  (  x      /    2  ), but I have no idea how to do it for   k  ≥  1.

tilsjaskak6 · Accepted Answer

If contour integration is acceptable, let   α  =            1      −              1        −        4                  τ          2                            2      τ       and   1      /    α  =            1      +              1        −        4                  τ          2                            2      τ      .                    (1)                              ∫          0          π                                      cos            ⁡            (            k            x            )                                1            −            2            τ            cos            ⁡            (            x            )                                            d                x                            =                  1          2                          ∫                      −            π                    π                                      cos            ⁡            (            k            x            )            +            i            sin            ⁡            (            k            x            )                                1            −            2            τ            cos            ⁡            (            x            )                                            d                x                            (2)                                  =                  1          2                          ∮                                    |                        z                          |                        =            1                                                z            k                                1            −            τ            z            −            τ                          z                              −                1                                                                                                d                        z                                i            z                                              (3)                                  =                  1                      2            i                          ∮                              z            k                                z            −            τ                          z              2                        −            τ                                            d                z                            (4)                                  =                  i                      2            τ                          ∮                              z            k                                (            z            −            α            )            (            z            −            1                          /                        α            )                                            d                z                            (5)                                  =                  1                      2            i                          1              −              4                              τ                2                                                    ∮                  (                                    z              k                                      z              −              α                                −                                    z              k                                      z              −              1                              /                            α                                )                          d                z                            (6)                                  =                              π                          α              k                                            1            −            4                          τ              2                                          Explanation:(1): since cosine is even, we can double the domain and divide by 2 since sine is odd, adding it to the integral has no effect(2): write trig functions as functions of z on the unit circle (  z  =      e          i      x        )(3): simplify(4): factor the denominator(5): partial fractions(6):   1      /    α is outside the unit circle and       Res          z      =      α        ⁡      (                  z        k                    z        −        α              )    =      α    k

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