An integral involving the inverse of f ( x ) = log ⁡ x − log ⁡ cos ⁡ x + x tan ⁡ xLet the function f : ( 0 , π 2 ) → R be defined as f ( x ) = log ⁡ x − log ⁡ cos ⁡ x + x tan ⁡ x .Let its inverse be denoted as f ( − 1 ) : R → ( 0 , π 2 ) , it satisfies ∀ z ∈ R , f ( f ( − 1 ) ( z ) ) = z .Consider the integral I = ∫ 0 1 z e z f ( − 1 ) ( z ) d z .Is it possible to evaluate the integral I in a closed form?

Question

An integral involving the inverse of   f  (  x  )  =  log  ⁡  x  −  log  ⁡  cos  ⁡  x  +  x  tan  ⁡  xLet the function   f  :      (    0    ,                      π        2              )    →      R   be defined as  f  (  x  )  =  log  ⁡  x  −  log  ⁡  cos  ⁡  x  +  x  tan  ⁡  x  .Let its inverse be denoted as       f          (      −      1      )        :      R    →      (    0    ,                      π        2              )  , it satisfies   ∀  z  ∈      R    ,        f  (      f          (      −      1      )        (  z  )  )  =  z  .Consider the integral  I  =      ∫    0    1    z        e    z          f          (      −      1      )        (  z  )    d  z  .Is it possible to evaluate the integral   I in a closed form?

Antonio Dickerson · Accepted Answer

I know it has been some time now but we can transform this question into something more manageable. We&#039;ll just use the substitution   z  =  f  (  x  ) with   d  z  =      f    ′    (  x  )  d  x. The bounds of integration change to the points x where   f  (  x  ) is respectively 0 or 1. Call these points       x    0   and       x    1  . Numerically they are       x    0    =  0.576412723  …      x    1    =  0.807626251  …And so we have  I  =      ∫                  x        0                            x        1              f  (  x  )      e          f      (      x      )            f          (      −      1      )        (  f  (  x  )  )      f    ′    (  x  )  d  x  =      ∫                  x        0                            x        1              x      f    ′    (  x  )  f  (  x  )      e          f      (      x      )        d  xNow we perform integration by parts with  u  =  x  f  (  x  )  ,    d  u  =  (  f  (  x  )  +  x      f    ′    (  x  )  )  d  x  d  v  =      f    ′    (  x  )      e          f      (      x      )        d  x  ,    v  =      e          f      (      x      )      and the integral becomes  I  =      x    1    f  (      x    1    )      e          f      (              x        1            )        −      x    0    f  (      x    0    )      e          f      (              x        0            )        −      ∫                  x        0                            x        1              f  (  x  )      e          f      (      x      )        d  x  −      ∫                  x        0                            x        1              x      f    ′    (  x  )      e          f      (      x      )        d  xSince    f  (      x    0    )  =  0 and   f  (      x    1    )  =  1  we have  I  =      x    1    −      ∫                  x        0                            x        1              f  (  x  )      e          f      (      x      )        d  x  −      ∫                  x        0                            x        1              x      f    ′    (  x  )      e          f      (      x      )        d  xNow we perform integration by parts on the latter integral with  u  =  x  ,    d  u  =  d  x  d  v  =      f    ′    (  x  )      e          f      (      x      )        d  x  ,    v  =      e          f      (      x      )      and we get  I  =      x    1    −      x    1        e          f      (              x        1            )        +      x    0        e          f      (              x        0            )        −      ∫                  x        0                            x        1              f  (  x  )      e          f      (      x      )        d  x  +      ∫                  x        0                            x        1                  e          f      (      x      )        d  xIt turns out we can perform the integration in the first integral since      e          f      (      x      )        =      x          cos      ⁡      x            e          x      tan      ⁡      x      and  ∫      e          f      (      x      )        d  x  =      e          x      tan      ⁡      x        cos  ⁡  xAfter cleaning up a bit  I  =      x    0    +      e                  x        1            tan      ⁡              x        1              cos  ⁡      x    1    −      e                  x        0            tan      ⁡              x        0              cos  ⁡      x    0    −      ∫                  x        0                            x        1              f  (  x  )      e          f      (      x      )        d  xand this is where I am getting stuck. Mathematica is having a problem with it too. On   [      x    0    ,      x    1    ]  we have  f  (  x  )      e          f      (      x      )        =            x              e                  x          tan          ⁡          (          x          )                            cos      ⁡      x            (    ln    ⁡          (              x                  cos          ⁡          x                    )        +    x    tan    ⁡    x    )  We also have another form  f  (  x  )      e          f      (      x      )        =      (                            x          2                sin        ⁡        x                              cos          2                ⁡        x              +          x              cos        ⁡        x              ln    ⁡          (              x                  cos          ⁡          x                    )        )        e          x      tan      ⁡      x      which suggest that we might have better luck looking at the following two integrals  J  =      ∫                  x        0                            x        1                                x        2            sin      ⁡      x                      cos        2            ⁡      x            e          x      tan      ⁡      x        d  xand  K  =      ∫                  x        0                            x        1              ln  ⁡      (          x              cos        ⁡        x              )        e          x      tan      ⁡      x        d  xThe other problem perhaps, is that       x    0   and       x    1   are given numerically. I have no idea if one could find a closed for expression for them or even perhaps something like a series representation.EDIT1: I&#039;m not sure if this could be called a reduction but using the equation defining   f  (  x  ) we can say that  cos  ⁡      x    1        e                  x        1            tan      ⁡              x        1              =      x    1        e          −      1      +      2              x        1            tan      ⁡              x        1              cos  ⁡      x    1    0      e                  x        0            tan      ⁡              x        0              =      x    0        e          2              x        0            tan      ⁡              x        0

An integral involving the inverse of f ( x ) = log ⁡ x

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