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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 http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=calculus&s2=integrate&s3=advanced [Broken] 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).
    Last edited by a moderator: May 2, 2017
  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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