Quick Integral

1. Apr 18, 2006

bomba923

(This is not homework, just part of a curious problem)

Is there an analytical (non-numerical) method/way to evaluate
$$\int\limits_0^{2\pi } {\sqrt {1 + \left( {h k\cos kx} \right)^2 } dx}$$
?

Last edited: Apr 18, 2006
2. Apr 19, 2006

bomba923

Alright, perhaps I should explain a bit...
(the "curious problem" I referred to)

Let $$y=h\sin kx$$, where $h,k \in \mathbb{R}^{+}$.

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,
$$L\left( {h,k} \right) = \int\limits_0^{2\pi } {\sqrt {1 + \left( {hk\cos kx} \right)^2 } dx}$$

Obviously,
$$\frac{\partial L}{\partial h} > 0\;{\text{and }}\frac{\partial L}{\partial k} > 0$$
--------------------
*But...precisely 'how' does $L$ increase with $h$ and/or $k$ ?

Is
$$\frac{{\partial ^2 L}}{{\partial h^2 }} > 0\;{\text{and }}\frac{{\partial ^2 L}}{{\partial k^2 }} > 0\;?$$

Also, is
$$\frac{{\partial L}}{{\partial h}} > \frac{{\partial L}}{{\partial k}}\;?$$

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

so that I may derive ∂L/∂h and ∂L/∂k, as well as ∂L2/∂h2 and ∂L2/∂k2 :shy:

Last edited: Apr 19, 2006
3. Apr 19, 2006

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 [Broken] 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).

Last edited by a moderator: May 2, 2017
4. Apr 20, 2006

bomba923

When I simplified the integrator's evaluation (remember that $h,k \in \mathbb{R}^{+}$),
I find that
$$L\left( {h,k} \right) = \int\limits_0^{2\pi } {\sqrt {1 + \left( {hk\cos 2k\pi} \right)^2 } dx} = \frac{{\sqrt {h^2 k^2 + 1} }}{k}E\left( {2k\pi\left| {\frac{{h^2 k^2 }}{{h^2 k^2 + 1}}} \right.} \right)$$

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

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

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

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

?

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

dextercioby

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

Daniel.