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A challenging Integral

  1. May 22, 2006 #1

    benorin

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    Put

    [tex]I_n(x)=\int \frac{n!dx}{x(x+1)\cdots (x+n)}[/tex]​

    The desired result is an evaluation of the sum

    [tex]S_n(x)=\sum_{k=0}^{n}(-1)^{k}\left(\begin{array}{c}n\\k\end{array}\right) (x+k)\log (x+k)[/tex]​

    and so it is important that the evaluation of [tex]I_n(x)[/tex] be not by means of partial fraction decomposition, for this is already of note that [tex]S_n(x)[/tex] may be evaluated, that is to say: it is known to me that

    [tex]\int\frac{n!dx}{x(x+1)\cdots (x+n)}=\int\sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array}\right) \frac{(-1)^k}{x+k}dx = \sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array}\right) (-1)^{k}\log (x+k)+C[/tex]​

    whence integrating once again will produce an evaluation for [tex]S_n(x)[/tex].

    I was working on developing a recurrence relation for [tex]I_n(x)[/tex] that I may solve it via generating functions; here of note is that it is also known to me that

    [tex]I_{n+1}(x)=I_n(x)-I_n(x+1)[/tex]

    yet this is not useful to that end. What is desired is a non-homogeneous recurrence. Any thoughts?
     
    Last edited: May 22, 2006
  2. jcsd
  3. May 22, 2006 #2

    Hurkyl

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    How does S relate to I?

    Is there any reason why you can't use the partial fraction decomposition, and then rearrage it into the form you desire?
     
  4. May 22, 2006 #3

    benorin

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    I made a mistake in my tex, now the three dots show up.
     
  5. May 22, 2006 #4

    benorin

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    To answer your question,

    [tex]\int I_n(x)dx = S_n(x) +g(x)[/tex]

    where I suspect that g(x) is rather tivial.

    EDIT: in fact, g(x)=0.
     
    Last edited: May 22, 2006
  6. May 22, 2006 #5

    nrqed

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    You don't really mean the integral of I_n(x), right? You mean I_n(x) = S_n(x), correct?

    EDIT: Actually, I am a bit confused by the notation.I is an integral over x so why is it a function of x? Fo you mean [itex] I_n(x) = \int_0^x \, f(x') [/itex] or something to that effect?
     
    Last edited: May 22, 2006
  7. May 22, 2006 #6

    nrqed

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    I haven't thought this one through completely but have you heard of Feynman's trick for combining propagators in quantum field theory. It *might* prove useful. It is the identity

    [tex] { 1 \over A_1 A_2 \ldots A_n} = \int_0^1 dy_1 \ldots dy_n \, \delta (\sum_{i=1}^n y_i -1) \, { (n-1)! \over (y_1 A_1 + y_2 A_2 + \ldots y_n A_n)^n } [/tex]

    I just have a *gut* feeling that this might do the trick but I haven't worked out anything. It's 1 Am here and I am about to go to bed.

    Patrick
     
  8. May 23, 2006 #7

    benorin

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    Because it is an indefinite integral, it is still a function of x after the integration is carried out, as in:[tex]\int \cos xdx=\sin x +C[/tex]
     
  9. May 23, 2006 #8

    benorin

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    And no, I do mean to say that [tex]\int I_n(x)dx = S_n(x) +C[/tex] since

    [tex]I_n(x)=\int\frac{n!dx}{x(x+1)\cdots (x+n)}=\int\sum_{k=0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right) \frac{(-1)^k}{x+k}dx = \sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array }\right) (-1)^{k}\log (x+k)+C_1[/tex]

    we have

    [tex]\int I_n(x)dx=\int\int\frac{n!dxdx}{x(x+1)\cdots (x+n)}= \int\sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array }\right) (-1)^{k}\log (x+k) dx +\int C_1 dx[/tex]
    [tex]= \sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array }\right) (-1)^{k}\int\log (x+k) dx+C_1x = \sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array }\right) (-1)^{k}\left[ (x+k)\log (x+k)-(x+k)\right] +C_1x+C_2[/tex]
    [tex]=\sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array }\right) (-1)^{k}(x+k)\log (x+k)-\sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array }\right) (-1)^{k}(x+k)+C_1x+C_2+h_n(x)=S_n(x)+C[/tex]

    where a formula from the calculus of finite differences has been used to determine that

    [tex]h_n(x)=\sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array }\right) (-1)^{k}(x+k)=\left\{\begin{array}{cc}x,& \mbox{ if } n=0\\-1, & \mbox{ if } n=1\\0, & \mbox{ if } n\geq 1 \end{array}\right.[/tex]​
     
    Last edited: May 24, 2006
  10. May 23, 2006 #9

    benorin

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    An interesting development

    An interesting development: recall that

    [tex]\log x = \int_{0}^{\infty}\left( e^{-t}-e^{-xt}\right)\frac{dt}{t}[/tex]​

    so that (dropping the "+C") we have

    [tex]I_n(x+1)=\sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array}\right) (-1)^{k}\log (x+k+1)=\sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array}\right) (-1)^{k}\int_{0}^{\infty}\left( e^{-t}-e^{-(x+k+1)t}\right)\frac{dt}{t} [/tex]
    [tex]= \int_{0}^{\infty}e^{-t}\sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array}\right) (-1)^{k}\left( 1-e^{-(x+k)t}\right)\frac{dt}{t}=\int_{0}^{\infty}\left[e^{-t}\delta_{n,0}- e^{-t(x+1)}\sum_{k=0}^{n}\left(\begin{array}{c}n\\k\end{array}\right) \left(-e^{-t}\right) ^{k}\right] \frac{dt}{t}[/tex]
    [tex]=\int_{0}^{\infty}\left[e^{-t}\delta_{n,0}- e^{-t(x+1)}\left(1-e^{-t}\right) ^{n}\right] \frac{dt}{t}[/tex]​

    where [tex]\delta_{n,0} = \left\{\begin{array}{cc}1,&\mbox{ if }
    n=0\\0, & \mbox{ if } n\neq 0\end{array}\right.[/tex] is the discete delta function, hence the evaluation

    [tex]I_n(x)=\left\{\begin{array}{cc}\int_{0}^{\infty}\left(e^{-t}- e^{-xt}\right) \frac{dt}{t}=\log x ,&\mbox{ if }
    n=0\\ -\int_{0}^{\infty}e^{-tx}\left(1-e^{-t}\right) ^{n}\frac{dt}{t}, & \mbox{ if } n\neq 0\end{array}\right.[/tex]​
     
    Last edited: May 24, 2006
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