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Integral evaluation - analytical vs. numerical

  1. May 19, 2005 #1
    Hi,

    Does anyone know a reason why [tex]\int_{-\infty}^{\infty}\cosh(x)^{-n}dx[/tex] (n>0) can be evaluated analytically when n = 1,2,3,..., but only numerically when n is non-integer. I don't know if there is a "reason", but I'm using this result in a Quantum Mechanics project and it would be cool if I could give some kind of intuitive reason why this is so.
     
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  3. May 19, 2005 #2

    arildno

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    My best guess:
    You may use the residue theorem from complex analysis to gain an exact expression in the case of integer values, whereas this won't work if you have a non-integer.
    I might be wrong, though..
     
  4. May 19, 2005 #3

    dextercioby

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    I've attached the antiderivative and here's your integral

    [tex] \int_{-\infty}^{+\infty} \frac{dx}{\cosh^{n} x}=\frac{\sqrt{\pi}}{2}\frac{\Gamma \left(\frac{n}{2}\right)}{\Gamma \left(\frac{n+1}{2}\right)} [/tex]

    "n" can be complex,even.

    Daniel.

    EDIT:The correct result is multiplied by 2.See posts #4 & #7.
     

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    Last edited: May 19, 2005
  5. May 19, 2005 #4

    Zurtex

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    O.K, in Mathematica I go that if [itex]\Re (n) > 0[/itex] then:

    [tex] \int_{-\infty}^{+\infty} \cosh^{-n} (x) dx = \frac{ \sqrt{\pi} \, \Gamma\left( \frac{n}{2} \right)}{ \Gamma\left( \frac{n + 1}{2} \right)}[/tex]
     
  6. May 19, 2005 #5

    dextercioby

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    How do you explain the "1/2" factor discrepancy...?

    Daniel.

    EDIT:See post #7 for details.
     
    Last edited: May 19, 2005
  7. May 19, 2005 #6

    Zurtex

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    Well if I left n=1 and integrate in mathematica I get [itex]\pi[/itex] and:

    [tex]\Gamma \left( \frac{1}{2} \right) = \sqrt{\pi}[/tex]

    Are you sure you did not integrate between 0 and Infinity?
     
    Last edited: May 19, 2005
  8. May 19, 2005 #7

    dextercioby

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    I got it,i plugged only for half a domain

    Using [1]

    [tex] \int_{0}^{+\infty} \frac{\sinh^{\mu}x}{\cosh^{\nu}x} \ dx =\frac{1}{2}B\left(\frac{\mu +1}{2},\frac{\nu-\mu}{2}\right) [/tex] (1)

    ,provided that

    [tex] \mbox{Re} \left(\mu\right) >-1 \ ; \ \mbox{Re} \left(\mu-\nu\right) <0 [/tex] (2)

    Making [itex] \mu=0 [/itex] (3) in (1) (admissible according to (2) and implying the condition [itex] \mbox{Re} (\nu) >0 [/itex] (4) ),one gets

    [tex] \int_{0}^{\infty} \frac{dx}{\cosh^{\nu} x} \ dx =\frac{1}{2} B\left(\frac{1}{2},\frac{\nu}{2}\right)=\frac{1}{2}\frac{\Gamma \left(\frac{1}{2}\right)\Gamma \left(\frac{\nu}{2}\right)}{\Gamma\left(\frac{\nu+1}{2}\right)} [/tex] (5)

    Using

    [tex] \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi} [/tex] (6)

    and

    [tex] \int_{\mathbb{R}} \frac{dx}{\cosh^{\nu} x} \ dx =2 \int_{0}^{\infty} \frac{dx}{\cosh^{\nu} x} \ dx [/tex] (7)

    ,one gets exactly the formula posted by Zurtex.

    Daniel.

    --------------------------------------------------
    [1]G & R,5-th edition,Academic Press,CD version,1996,page 388,formula 3.512-2.
     
  9. May 19, 2005 #8

    dextercioby

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    Yes,i used an online integrator and indeed integrated only half of the domain.

    I'll edit and make a reference to your post & mine just above.

    Daniel.
     
  10. May 22, 2005 #9

    uart

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    So just in case anyone is at all unclear about how this relates to the original observation that :
    It's because [tex]\Gamma(x)[/tex] is easily expressable in terms of elementary functions and constants for whole and half integers but generally needs other forms of numerical calulation for arbitrary values of "x".
     
    Last edited: May 22, 2005
  11. May 22, 2005 #10

    dextercioby

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    Incidentally,both beta & gamma-Euler are tabulated for a certain range of values of their arguments.

    So the integral is solvable analytically.Expressing its values in terms of analytical special functions means just that.

    Daniel.
     
  12. May 22, 2005 #11

    krab

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    That means that whether an integral has an analytical solution or not is based on convention (how many special functions are named). I don't find such a distinction very useful. I would echo uart's comment.
     
  13. May 25, 2005 #12
    Thanks y'all. I've included a bit about the gamma function, interesting stuff...
     
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