How do you show that ∫ 0 ∞ log ⁡ | tan ⁡ x | 1 + x 2 d x = π 2 log ⁡ ( tanh ⁡ 1 ) ?

Question

How do you show that      ∫          0              ∞                  log      ⁡              |            tan      ⁡      x              |                    1      +              x                  2                        d  x  =      π    2    log  ⁡  (  tanh  ⁡  1  )    ?

cokeman206 · Accepted Answer

How about:                              (                      ∗                    )                                      ∫          0          ∞                                      log            ⁡                          |                        tan            ⁡            x                          |                                            1            +                          x              2                                                d                x        =                  1          2                          ∫          0          ∞                                      log            ⁡                          tan              2                        ⁡            x                                1            +                          x              2                                                d                x        =                  1          4                          ∫                      −            ∞                    ∞                                      log            ⁡                          tan              2                        ⁡            x                                1            +                          x              2                                                d                x            Writing the integral over       R   as the average of integrals over       R    +  i  ϵ and       R    −  i  ϵ and using   |  tan  ⁡  (  z  )      |    2    =                    sin        2            ⁡      (      2      x      )      +              sinh        2            ⁡      (      2      y      )              (      cos      ⁡      (      2      x      )      +      cosh      ⁡      (      2      y      )              )        2                  →          y      →      ±      ∞        1 we can complete the integration contours and apply the residue theorem:                              ∫                      −            ∞                    ∞                                      log            ⁡                          tan              2                        ⁡            x                                1            +                          x              2                                                d                x                            =                              1          2                          ∫                      −            ∞            +            i            0                                ∞            +            i            0                                                log            ⁡                          tan              2                        ⁡            x                                1            +                          x              2                                                d                x        +                  1          2                          ∫                      −            ∞            −            i            0                                ∞            −            i            0                                                log            ⁡                          tan              2                        ⁡            x                                1            +                          x              2                                                d                x                                          =                              1          2                ⋅        2        π        i                  Res                      x            =            i                          ⁡                              log            ⁡                          tan              2                        ⁡            x                                1            +                          x              2                                      −                  1          2                ⋅        2        π        i                  Res                      x            =            −            i                          ⁡                              log            ⁡                          tan              2                        ⁡            x                                1            +                          x              2                                                                        =                    2        π        log        ⁡        tanh        ⁡        (        1        )            Combining with eq.      ∫    0    ∞              log      ⁡              |            tan      ⁡      x              |                    1      +              x        2                  d    x  =      π    2    log  ⁡  tanh  ⁡  (  1  )

robbbiehu · Accepted Answer

Denote the evaluated integral as I, then I may be rewritten as  I  =      1    2        ∫    0    ∞              ln      ⁡              sin        2            ⁡      x              1      +              x        2                d  x  −      1    2        ∫    0    ∞              ln      ⁡              cos        2            ⁡      x              1      +              x        2                d  xUsing Fourier series representations of   ln  ⁡      sin    2    ⁡  θ and   ln  ⁡      cos    2    ⁡  θand  ln  ⁡      cos    2    ⁡  θ  =  −  2  ln  ⁡  2  +  2      ∑          k      =      1        ∞    (  −  1      )          k      +      1                  cos      ⁡      2      k      θ        k  also note that      ∫    0    ∞              cos      ⁡      a      x              1      +              x        2                d  x  =            π              e                  −          a                      2  then                    I                            =        −                  ∑                      k            =            1                    ∞                          1          k                          ∫          0          ∞                                      cos            ⁡            2            k            x                                1            +                          x              2                                              d        x        −                  ∑                      k            =            1                    ∞                                      (            −            1                          )                              k                +                1                                              k                          ∫          0          ∞                                      cos            ⁡            2            k            x                                1            +                          x              2                                              d        x                                          =        −        π                  ∑                      k            =            1                    ∞                                      e                          −              2              k                                            2            k                          −        π                  ∑                      k            =            1                    ∞                (        −        1                  )                      k            +            1                                                e                          −              2              k                                            2            k                                                            =                  π          2                ln        ⁡                  (          1          −                      e                          −              2                                )                −                  π          2                ln        ⁡                  (          1          +                      e                          −              2                                )                                                  =                  π          2                ln        ⁡                  (                                    1              −                              e                                  −                  2                                                                    1              +                              e                                  −                  2                                                              )                                                  =                  π          2                ln        ⁡                  (          tanh          ⁡          1          )

How do you show that int_0^infty (log|tan x|)/(1+x^2)dx=pi/2log(tanh 1) ?

Answered question

Answer & Explanation

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