An Analytical Approach to Evaluating an Integral

[bomba923]

(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]
?

 

[bomba923]

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]

Obviously,
[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] ?

Is
[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:

 

[benorin]

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 and put

sqrt(1+(h*k*cos(k*x))^2)

into the input box marked EXPRESSION, and put

x,0,2pi

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).

 

[bomba923]

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]

and
[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]

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

?

 

[dextercioby]

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

Daniel.