Question releating to the ∫ 1 x ln ⁡ ( t ) t + 1 If f ( x ) = ∫ 1 x ln ⁡ ( t ) t + 1 d t if x &gt; 0. Compute f ( x ) + f ( 1 / x ). As a check, you should obtain f ( 2 ) + f ( 1 / 2 ) = ( ln ⁡ 2 ) 2 I have tried evaluating the integral ∫ 1 x ln ⁡ ( t ) t + 1 d t by parts, but it turns out to be futile as ∫ ln ⁡ ( t ) t + 1 d t = ln ⁡ ( t ) ln ⁡ ( t + 1 ) − ∫ ln ⁡ ( t + 1 ) t t and ∫ ln ⁡ ( t + 1 ) t = ln ⁡ ( t ) ln ⁡ ( t + 1 ) − ∫ ln ⁡ ( t ) t + 1 d t which leads to ∫ ln ⁡ ( t ) t + 1 d t = ∫ ln ⁡ ( t ) t + 1 d tIs there any other way to evaluate this integral?

Question

Question releating to the       ∫    1    x              ln      ⁡      (      t      )              t      +      1      If   f  (  x  )  =      ∫    1    x              ln      ⁡      (      t      )              t      +      1        d  t if   x  &amp;gt;  0. Compute   f  (  x  )  +  f  (  1      /    x  ). As a check, you should obtain   f  (  2  )  +  f  (  1      /    2  )  =  (  ln  ⁡  2      )    2  I have tried evaluating the integral       ∫    1    x              ln      ⁡      (      t      )              t      +      1        d  t by parts, but it turns out to be futile as   ∫            ln      ⁡      (      t      )              t      +      1        d  t  =  ln  ⁡  (  t  )  ln  ⁡  (  t  +  1  )  −  ∫            ln      ⁡      (      t      +      1      )        t    t and   ∫            ln      ⁡      (      t      +      1      )        t    =  ln  ⁡  (  t  )  ln  ⁡  (  t  +  1  )  −  ∫            ln      ⁡      (      t      )              t      +      1        d  t which leads to   ∫            ln      ⁡      (      t      )              t      +      1        d  t  =  ∫            ln      ⁡      (      t      )              t      +      1        d  tIs there any other way to evaluate this integral?

Nancy Phillips · Accepted Answer

You do not have to evaluate the integral. Note that                    f        (        1                  /                x        )                            =                  ∫          1                      1                          /                        x                                                log            ⁡            t                                t            +            1                                  d        t                                          =                  ∫          1          x                                      log            ⁡            1                          /                        τ                                              1              τ                        +            1                                            (          −                      1                          τ              2                                )                d        τ                              substition                       t            =                          1              τ                                                                        =                  ∫          1          x                                      log            ⁡            τ                                τ            +                          τ              2                                              d        τ                                          =                  ∫          1          x                                      log            ⁡            τ                    τ                        d        τ        −                  ∫          1          x                                      log            ⁡            τ                                1            +            τ                                  d        τ                                          =                  ∫          0                      log            ⁡            x                          u                d        u        −        f        (        x        )                                          =                  1          2                          log          2                ⁡        x        −        f        (        x        )        .

gobeurzb · Accepted Answer

What you want is  f  (  x  )  +  f  (  1      /    x  )thus      ∫    1    x              ln      ⁡      (      t      )              1      +      t        d  t  +      ∫    1          1              /            x                  ln      ⁡      (      t      )              t      +      1        d  twhich can be written as      ∫    1    x              ln      ⁡      (      t      )              1      +      t        d  t  +      ∫    1    x              ln      ⁡      (      1              /            t      )              1              /            t      +      1        d  [  1      /    t  ]Thus  f  (  x  )  +  f  (  1      /    x  )  =      ∫    1    x              ln      ⁡      (      t      )        t    d  tAnd this can be calculated  ∫            ln      ⁡      (      t      )        t    d  t  =                    ln        2            ⁡      (      t      )        2  So you find  f  (  x  )  +  f  (  1      /    x  )  =                    ln        2            ⁡      (      x      )        2

If f(x)=int_1^x (ln(t))/(t+1)dt if x>0. Compute f(x)+f(1/x). As a check, you should obtain f(2)+f(1/2)=(ln2)^2

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