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Prove that for n> 1 integral

  1. Nov 7, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove that for n> 1
    [tex]
    \int\limits_{0}^{\infty}{\frac{1}{(x + \sqrt{x^2 + 1})^n}} \mathrm{d}x = \frac{n}{n^2 - 1}
    [/tex]
    2. Relevant equations



    3. The attempt at a solution

    Tried substitute x = cosh theta, then

    [tex]
    \frac{\mathrm{dx}}{\mathrm{d}\theta} = \sinh \theta
    [/tex]

    [tex]
    \int\limits_{0}^{\infty}{\frac{1}{(x + \sqrt{x^2 + 1})^n}} \mathrm{d}x = \int\limits_{0}^{\infty}{\frac{\sinh{\theta}}{(\cosh{\theta} + \sinh{\theta})^n }
    [/tex]
    I'm getting in the right direction here? I'm really stuck..
     
    Last edited: Nov 7, 2008
  2. jcsd
  3. Nov 7, 2008 #2

    HallsofIvy

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    Re: Integral

    Isn't something missing here?
     
  4. Nov 7, 2008 #3
    Re: Integral

    n > 1
     
  5. Nov 7, 2008 #4

    gabbagabbahey

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    Re: Integral

    I think he meant your attempt at a solution :wink:
     
  6. Nov 7, 2008 #5
    Re: Integral

    well I believe that mentioning n > 1 also is quite important... I've tried to use sinh-substitution, but can't really say I'm getting any wiser about it.
     
  7. Nov 7, 2008 #6

    Dick

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    Re: Integral

    Use cosh(x)=(e^x+e^(-x))/2 and sinh(x)=(e^x-e^(-x))/2. Now try to integrate it. It's really pretty easy.
     
  8. Nov 7, 2008 #7
    Re: Integral

    [tex]

    \int\limits_{0}^{\infty}{\frac{1}{(x + \sqrt{x^2 + 1})^n}} \mathrm{d}x = \int\limits_{x = 0}^{x = \infty}{\frac{\sinh{\theta}}{(\cosh{\theta} + \sinh{\theta})^n } \mathrm{d}\theta

    [/tex]

    [tex]
    = \int\limits_{x = 0}^{x = \infty} \frac{e^{\theta} - e^{-\theta}}{(e^\theta)^n} \mathrm{d}\theta
    [/tex]
    Don't know what to do here.
     
    Last edited: Nov 8, 2008
  9. Nov 7, 2008 #8

    Dick

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    Re: Integral

    Split it into two integrals. Use rules of exponents. And you are missing a factor of 1/2, I think.
     
  10. Nov 8, 2008 #9
    Re: Integral

    [tex]
    \frac{1}{2} \int\limits_{x = 0}^{x = \infty} {e^{\theta - n\theta}} \mathrm{d}\theta -
    \frac{1}{2} \int\limits_{x = 0}^{x = \infty}{e^{-\theta - n\theta} \mathrm{d}\theta =


    [/tex]

    Then integrating and substitute back, don't really seem to get rid of those exponentials..
     
  11. Nov 8, 2008 #10

    Dick

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    Re: Integral

    Just continue. Write down the antiderivatives. n>1. The contribution from infinity vanishes. It's the theta=0 part that counts. And e^0=1. See, no exponential?
     
  12. Nov 8, 2008 #11

    HallsofIvy

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    Re: Integral

    Your original substitution x= cosh(theta) has a fundamental problem: the lower limit of integration is x= 0 and cosh(theta) is never 0. I would suggest x= tan(y) instead.
     
  13. Nov 8, 2008 #12
    Re: Integral

    Is that a problem? when integrating and substituting back the cosh dissappears does'nt it?
     
  14. Nov 8, 2008 #13

    Dick

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    Re: Integral

    Yeah, it's a problem. If you use x=cosh(theta), 1+cosh(theta)^2 isn't equal to sinh(theta)^2. It's the other way around. Use x=sinh(theta). (You had said you were using sinh-substitution, and I didn't notice you had it backwards.) It looks almost the same as x=cosh(theta) except for a sign difference.
     
    Last edited: Nov 8, 2008
  15. Nov 8, 2008 #14
    Re: Integral

    So
    [tex]
    \int\limits_{0}^{\intfty}{\frac{\cosh{\theta}}{(\sinh{\theta} + \cosh{\theta})^n}}\mathrm{d}\theta = \frac{1}{2}\int\limits_{x = 0}^{x = \infty}{\frac{e^{x} + e^{-x}}{(e^x)^n}}{\mathrm{d}\theta}
    [/tex]
    and

    [tex]
    \frac{1}{2}\int\limits_{x = 0}^{x = \infty}{e^{\theta - n\theta}}\mathrm{d}\theta + \frac{1}{2}\int\limits_{x = 0}^{x = \infty}{e^{-\theta - n\theta}}\mathrm{d}\theta

    [/tex]

    and then factorize e^{1 - n} out?
    [tex]
    \frac{e^{1 - n}}{2} \int\limits_{x = 0}^{x = \infty}{cosh \theta}~ \mathrm{d}\theta
    [/tex]
     
  16. Nov 8, 2008 #15

    Dick

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    Re: Integral

    Noooo. And you can change your limits to theta=0 etc, ok? Your first integral is e^((1-n)*theta). That's EASY to integrate.
     
  17. Nov 8, 2008 #16
    Re: Integral

    ahh. figured out, thanks
     
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