Integral ∫ 0 1 ln ⁡ x x 2 − 1 d xPlease help evaluating this integralMathematica could not evaluate it in a closed form. Numerically it is about I ≈ 1.060837861412045137097... ,

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Integral       ∫          0        1                      ln        ⁡                  x                                      x          2                −        1              d  xPlease help evaluating this integralMathematica could not evaluate it in a closed form. Numerically it is about  I  ≈  1.060837861412045137097...  ,

lywyk0 · Accepted Answer

Let&#039;s make a variable change   x  =      e          −      t        I  =      ∫    0          ∞                          t                    e                  −          t                            1      −              e                  −          2          t                      d  tNow, let&#039;s expand the integrand into Taylor series:      1          1      −              e                  −          2          t                      =      ∑          n      =      0              ∞                  (      2      n      )      !              (              2        n            n      !              )        2                  e          −      2      n      t      Thus  I  =      ∑          n      =      0              ∞                  (      2      n      )      !              (              2        n            n      !              )        2                  ∫    0          ∞            t        e          −      (      2      n      +      1      )      t        d  t  =  =            π        2        ∑          n      =      0              ∞                  (      2      n      )      !              (              2        n            n      !              )        2                  1          (      2      n      +      1              )                              3            2                              Now, by hand, if we take only the three first terms of the sum we get   I  =  1.001...Not bad!

historia075c · Accepted Answer

The Lerch transcendent, initially defined by  Φ  (  z  ,  s  ,  a  )  :=      ∑          k      =      0        ∞              z      k              (      a      +      k              )        s              ,    a  &amp;gt;  0  ,  ℜ  s  &amp;gt;  1  ,      |    z      |    &amp;lt;  1  ,admits the following integral representation  Φ  (  z  ,  s  ,  a  )  =      ∫    0          ∞                          x                  s          −          1                            e                  −          a          x                            1      −      z              e                  −          x                          d    x  .By differentiation      ∂    z    r    Φ  (  z  ,  s  ,  a  )  =  (  −  1      )    r        ∫    0          ∞                          x                  s          −          1                            e                  −          (          a          +          r          )          x                            (      1      −      z              e                  −          x                            )                  r          +          1                          d    x  ,then, by extension, your integral   I may be formally rewritten as  I  =      ∫          0        1                      ln        ⁡                  x                                      x          2                −        1              d  x  =  −  i      2    π          ∂    z                −              1        2              Φ      (    1    ,          3      2        ,    1    )  This is to show the level of complexity of this integral: fractional calculus.

Please help evaluating this integral int_(0)^1 sqrt((ln(x))/(x^2-1}} dx Mathematica could not evaluate it in a closed form. Numerically it is about I≈1.060837861412045137097..., ISC+ did not return any closed form for it.

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