Evaluate ∫ 0 ∞ ln ⁡ x ( x 2 + 1 ) n d x

Question

Evaluate       ∫          0              ∞                  ln      ⁡      x                      (                  x                      2                          +        1        )                    n              d  x

Ryan Fitzgerald · Accepted Answer

The derivative can be taken using the product rule:

\frac{d^{n}}{d a^{n}} \frac{\ln a}{\sqrt{a}} = \sum_{k = 0}^{n} (\binom{n}{k}) \frac{d^{k} a^{- 1 / 2}}{d a^{k}} \frac{d^{n - k} \ln a}{d a^{n - k}}

together with

\frac{d^{k} a^{λ}}{d a^{k}} = k! (\binom{λ}{k}) a^{λ - k}

and

\frac{d^{k} \ln a}{d a^{k}} = \frac{d^{k - 1} a^{- 1}}{d a^{k - 1}}

Another way to evaluate the given integral is as follows:

I_{n} = \int_{0}^{\infty} \frac{\ln x d x}{(x^{2} + 1)^{n}} = \frac{1}{4} F_{n}^{'} (\frac{1}{2}), F_{n} (α) = 2 \int_{0}^{\infty} \frac{x^{2 α - 1} d x}{(x^{2} + 1)^{n}},

and the last integral is evaluated using the beta function

F_{n} (α) = B (α, n - α) = \frac{1}{(n - 1)!} \frac{π}{\sin α π} \prod_{k = 1}^{n - 1} (k - α),

so that

F_{n}^{'} (α) / F_{n} (α)

is obtained easily, and we get

I_{n} = - \frac{π}{4} (\binom{n - 3 / 2}{n - 1}) \sum_{k = 1}^{n - 1} \frac{1}{k - 1 / 2} = - \frac{π}{2} \frac{(2 n - 3)!!}{(2 n - 2)!!} \sum_{k = 1}^{n - 1} \frac{1}{2 k - 1} .

landdenaw · Accepted Answer

Use   (      a    x        )    ′    =      a    x    ln  ⁡  a to evaluate                                          d                          n                                            d                          a                              n                                                                          (                                          ln                ⁡                a                                            a                                      )                                a            =            1                          =                                          ∂                          n                                            ∂                          a                              n                                                                          (                                          ∂                                  a                  x                                                            ∂                x                                                                                      |                                                            x                =                −                                  1                  2                                                      )                                a            =            1                          =                  ∂                      ∂            x                                                (                                                            ∂                                      n                                                                    a                  x                                                            ∂                                  a                                      n                                                                                                                        |                                                            a                =                1                                      )                                x            =            −                          1              2                                                          =                              d                      d            x                                                (                                    ∏                      k            =            1                                n                          (        x        −        k        +        1        )                                            )                                            x            =            −                          1              2                                      =                  ∏                      k            =            1                                n                          (                  1          2                −        k        )                  ∑                      j            =            1                                n                                    1                                    1              2                        −            j                                              =                            −                              (            −            1                          )              n                        (            2            n            −            1            )            !            !                                2                          n              −              1                                                ∑                      j            =            1                                n                                    1                      2            j            −            1                              Thus      ∫          0              ∞                  ln      ⁡      x                      (                  x                      2                          +        1        )                    n        +        1              d  x  =            (      −      1              )                  n                    π              4      n      !                  d              n                    d              a                  n                                (                        ln          ⁡          a                          a                    )              a      =      1        =  −            π      (      2      n      −      1      )      !      !                      2                  n          +          1                    n      !            ∑          j      =      1              n            1          2      j      −      1

Evaluate <msubsup> ∫ <mrow class="MJX-TeXAtom-ORD"> 0 </mrow>

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New Questions in Calculus 1