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Quick Integral

  1. Apr 18, 2006 #1
    (This is not homework, just part of a curious problem)

    Is there an analytical (non-numerical) method/way to evaluate
    [tex]\int\limits_0^{2\pi } {\sqrt {1 + \left( {h k\cos kx} \right)^2 } dx} [/tex]
    Last edited: Apr 18, 2006
  2. jcsd
  3. Apr 19, 2006 #2
    Alright, perhaps I should explain a bit...
    (the "curious problem" I referred to)

    Let [tex]y=h\sin kx[/tex], where [itex]h,k \in \mathbb{R}^{+}[/itex].

    The length 'L' of this function on the x-interval [0,2π] can be expressed
    as a function of 'h' and 'k'. In other words,
    [tex]L\left( {h,k} \right) = \int\limits_0^{2\pi } {\sqrt {1 + \left( {hk\cos kx} \right)^2 } dx} [/tex]

    [tex]\frac{\partial L}{\partial h} > 0\;{\text{and }}\frac{\partial L}{\partial k} > 0[/tex]
    *But...precisely 'how' does [itex]L[/itex] increase with [itex]h[/itex] and/or [itex]k[/itex] ?

    [tex]\frac{{\partial ^2 L}}{{\partial h^2 }} > 0\;{\text{and }}\frac{{\partial ^2 L}}{{\partial k^2 }} > 0\;? [/tex]

    Also, is
    [tex]\frac{{\partial L}}{{\partial h}} > \frac{{\partial L}}{{\partial k}}\;?[/tex]

    *And so, to answer these questions,
    it would greatly help to analytically evaluate the integral
    [tex]\int\limits_0^{2\pi } {\sqrt {1 + \left( {hk\cos kx} \right)^2 } dx}[/tex]

    so that I may derive ∂L/∂h and ∂L/∂k, as well as ∂L2/∂h2 and ∂L2/∂k2 :shy:
    Last edited: Apr 19, 2006
  4. Apr 19, 2006 #3


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    Homework Helper

    The analytical evaluation of that integral involves an elliptic integral of the second kind (you may go to this page QuickMath integrator and put


    into the input box marked EXPRESSION, and put


    into the input box marked VARIABLE(S)&LIMITS to get an exact pression for the integral); know that an elliptic integral of the second kind is not an elementary function (e.g. it's not pretty).
  5. Apr 20, 2006 #4
    When I simplified the integrator's evaluation (remember that [itex]h,k \in \mathbb{R}^{+}[/itex]),
    I find that
    [tex]L\left( {h,k} \right) = \int\limits_0^{2\pi } {\sqrt {1 + \left( {hk\cos 2k\pi} \right)^2 } dx} = \frac{{\sqrt {h^2 k^2 + 1} }}{k}E\left( {2k\pi\left| {\frac{{h^2 k^2 }}{{h^2 k^2 + 1}}} \right.} \right) [/tex]

    To find ∂L/∂h and ∂L/∂k, I applied the Product Rule:
    [tex] \begin{gathered}
    \frac{\partial }{{\partial h}}\left[ {\frac{{\sqrt {h^2 k^2 + 1} }}{k}E\left( {2k\pi \left| {\frac{{h^2 k^2 }}{{h^2 k^2 + 1}}} \right.} \right)} \right] = \hfill \\
    \frac{{hk}}{{\sqrt {h^2 k^2 + 1} }}E\left( {2k\pi \left| {\frac{{h^2 k^2 }}{{h^2 k^2 + 1}}} \right.} \right) + \frac{{\sqrt {h^2 k^2 + 1} }}{k}\frac{\partial }
    {{\partial h}}E\left( {2k\pi \left| {\frac{{h^2 k^2 }}{{h^2 k^2 + 1}}} \right.} \right) \hfill \\
    \end{gathered} [/tex]

    [tex] \begin{gathered}
    \frac{\partial }{{\partial k}}\left[ {\frac{{\sqrt {h^2 k^2 + 1} }}{k}E\left( {2k\pi \left| {\frac{{h^2 k^2 }}{{h^2 k^2 + 1}}} \right.} \right)} \right] = \hfill \\
    \frac{{ - 1}}{{k^2 \sqrt {h^2 k^2 + 1} }}E\left( {2k\pi \left| {\frac{{h^2 k^2 }}{{h^2 k^2 + 1}}} \right.} \right) + \frac{{\sqrt {h^2 k^2 + 1} }}{k}\frac{\partial }
    {{\partial k}}E\left( {2k\pi \left| {\frac{{h^2 k^2 }}{{h^2 k^2 + 1}}} \right.} \right) \hfill \\
    \end{gathered} [/tex]

    *But, how do you differentiate an elliptic integral?
    -In other words, how can I find
    [tex]\frac{\partial }{{\partial h}}E\left( {2k\pi\left| {\frac{{h^2 k^2 }}
    {{h^2 k^2 + 1}}} \right.} \right) [/tex]

    [tex]\frac{\partial }{{\partial k}}E\left( {2k\pi\left| {\frac{{h^2 k^2 }}
    {{h^2 k^2 + 1}}} \right.} \right) [/tex]

    Last edited: Apr 20, 2006
  6. Apr 20, 2006 #5


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    Homework Helper

    Do some documentation on mathworld.com or some book on special functions...

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