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Hyperbolic integral

  1. Jun 20, 2015 #1
    What exactly is the integral of ##\frac{1}{\sqrt{x^2 - 1}}##?
    I know that the derivative of ##\cosh^{-1}{x}## is ##\frac{1}{\sqrt{x^2 - 1}}##, but ##\cosh^{-1}{x}## is only defined for ##x \geq 1##, whereas ##\frac{1}{\sqrt{x^2 - 1}}## is defined for all ##|x| \geq 1##. How do I take that into account? Do I write:
    $$\int \frac{1}{\sqrt{x^2 - 1}} dx = \cosh^{-1}{|x|} + c$$
    UPDATE: I tried differentiating ##\cosh^{-1}{|x|}## as a piecewise function and I ended up with ##-\frac{1}{\sqrt{x^2 - 1}}## for ##x \leq 1##. Now I'm more confused.
    Should I write the integral as:
    $$\int \frac{1}{\sqrt{x^2 - 1}} dx = S(x) \cosh^{-1}{|x|}$$
    where ##S## is the signum function?
     
    Last edited: Jun 20, 2015
  2. jcsd
  3. Jun 20, 2015 #2

    Matternot

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    Consider a comparison to ln(x)

    notice how:
    ##\frac{d}{dx}\ln(x)## is the same as ##\frac{d}{dx}\ln(-x)##

    This is inconsistent with your ##\cosh^{-1}{|x|}##

    The idea that resolves this is that you can use the idea that ##\frac{1}{\sqrt{x^2 - 1}}## is an even function.
    This leads to your last line which sounds right to me.

    There could be some quirks using complex numbers... I'll try it and get back.

     
    Last edited: Jun 20, 2015
  4. Jun 20, 2015 #3

    Matternot

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    one idea is to consider ##\cosh(x) = cos(ix)##

    i.e.
    ##\cosh^{-1}(x) = -i cos^{-1}(x)##
    this leads to simply ##±\frac{1}{\sqrt{x^{2}-1}}##

    Most obvious solution would be your last line
     
    Last edited: Jun 20, 2015
  5. Jun 21, 2015 #4
    I think I figured it out. It's ##\ln|x + \sqrt{x^2 - 1}|##.
    Which is identical to the inverse hyperbolic cosine, except it's defined for both positive and negative ##x + \sqrt{x^2 - 1}##.
     
  6. Jun 21, 2015 #5

    Matternot

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    yeah, that's a much nicer way of writing it.
     
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