An Analytical Approach to Evaluating an Integral

In summary, the conversation discusses the analytical evaluation of the integral \int\limits_0^{2\pi } {\sqrt {1 + \left( {hk\cos kx} \right)^2 } dx}, where y=h\sin kx and L(h,k) represents the length of this function on the interval [0,2π]. The conversation also delves into finding the partial derivatives of L with respect to h and k, and how to differentiate an elliptic integral in order to do so. The conversation ends with a recommendation to consult resources such as mathworld.com or books on special functions for more information.
  • #1
bomba923
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(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]
?
 
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  • #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]

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:
 
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  • #3
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).
 
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  • #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]

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]

?
 
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  • #5
Do some documentation on mathworld.com or some book on special functions...

Daniel.
 

FAQ: An Analytical Approach to Evaluating an Integral

What is an analytical approach to evaluating an integral?

An analytical approach to evaluating an integral involves using mathematical techniques to find the exact value of an integral. This includes using integration rules, substitution, and other advanced methods to manipulate the integral into a simpler form that can be solved.

Why is it important to use an analytical approach to evaluate integrals?

Using an analytical approach allows for a more precise and accurate evaluation of integrals compared to numerical methods. It also allows for a deeper understanding of the mathematical concepts behind integrals and their applications in various fields such as physics, engineering, and economics.

What are the common techniques used in an analytical approach to evaluating integrals?

Some common techniques used in an analytical approach include the fundamental theorem of calculus, integration by substitution, integration by parts, and trigonometric substitution. These techniques can be combined and used in various combinations to evaluate different types of integrals.

Are there any limitations to using an analytical approach to evaluating integrals?

Yes, there can be limitations to using an analytical approach. Some integrals may not have closed-form solutions and may require numerical methods to evaluate. Additionally, some integrals may be very complex and require advanced techniques or software to solve.

How can I improve my skills in using an analytical approach to evaluating integrals?

The best way to improve your skills is through practice and studying various examples and techniques. Working through problems, either by hand or using software, can help you gain a better understanding of the concepts and improve your problem-solving skills. It is also helpful to seek guidance from a mentor or tutor if you are struggling with certain concepts or techniques.

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