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Thread: can this integral be solved analytically!

  1. #1
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    can this integral be solved analytically!

    Hello,

    i like to find an analytical solution for the following integral:

    integral from [0 to pi]

    {cos(2nθ)/ (x^2+ cos^2(θ))}dθ

    x is constant and n is integer

    Appreciate your help!

    mopen
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  2. #2
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    Re: can this integral be solved analytically!

    Quote Originally Posted by mopen View Post
    Hello,

    i like to find an analytical solution for the following integral:

    integral from [0 to pi]

    {cos(2nθ)/ (x^2+ cos^2(θ))}dθ<------ What you have written here does not match with the image you have attached. What integral do you want to find,

    $\displaystyle \int_{0}^{\pi}\frac{\cos(n\theta)}{x^2+\cos^{2} \theta}d \theta\mbox{ or }\int_{0}^{\pi}\frac{\cos(2n\theta)}{x^2+\cos^{2} \theta}d \theta$


    x is constant and n is integer

    Appreciate your help!

    mopen
    .
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  3. #3
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    Re: can this integral be solved analytically!

    Hello Sudharaka,

    i like to evaluate the second integral :
    integral from [0 to pi]

    {cos(2nθ)/ (x^2+ cos^2(θ))}dθ

    which is the same as the one on the image but with only even n's .

    Thank You

    Mopen
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: can this integral be solved analytically!

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    Re: can this integral be solved analytically!

    But what does it means, Seriesexpansion of the integral at :t =‡ 0?
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  6. #6
    MHF Contributor chisigma's Avatar
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    Re: can this integral be solved analytically!

    Quote Originally Posted by mopen View Post
    Hello,

    i like to find an analytical solution for the following integral:

    integral from [0 to pi]

    {cos(2nθ)/ (x^2+ cos^2(θ))}dθ

    x is constant and n is integer

    Appreciate your help!

    mopen
    There are two ways to [try] to solve the integral...

    a) use the variable transformation...

    $\displaystyle t= \tan \frac{\theta}{2} \implies \theta= 2 \tan^{-1} t\ ;\ d \theta= 2 \frac{d t}{1+t^{2}}\ ;\ \cos \theta= \frac{1-t^{2}}{1+t^{2}} $ (1)

    b) use the identity...

    $\displaystyle \cos \theta= \frac{z+z^{-1}}{2}_{z=e^{i \theta}$ (2)

    ... and integrate along the unit circle in the complex plane using the Cauchy integral formulas...

    Neither a) nor b) is particulary comfortable but... what do You prefer?...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  7. #7
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    Re: can this integral be solved analytically!

    i prefer "a" , but how to substitute the cos(2nθ) term?
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  8. #8
    MHF Contributor chisigma's Avatar
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    Re: can this integral be solved analytically!

    Quote Originally Posted by mopen View Post
    i prefer "a" , but how to substitute the cos(2nθ) term?
    You can use the identity...

    $\displaystyle \cos k x = \frac{1}{2}\ \{(2 \cos x)^{k} -k (2 \cos x)^{k-1} + \frac{k}{2}\ \binom{k-3}{1}\ (2 \cos x)^{k-4} - ...\}$ (1)

    ... and that is why a) is not 'very comfortable' ...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  9. #9
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    Re: can this integral be solved analytically!

    Ok , what about the second option how to deal with it, and would letting n=1 make the integration easier?
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  10. #10
    MHF Contributor chisigma's Avatar
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    Re: can this integral be solved analytically!

    For the second option for the moment You can keep the term n and the first step is to set $\displaystyle \gamma=2 \theta$ so that the integral becomes...

    $\displaystyle I(x)= \frac{1}{2}\ \int_{0}^{2 \pi} \frac{\cos n \gamma}{x^{2} + \cos^{2} \frac{\gamma}{2}}\ d \gamma$ (1)

    ... that has limits $\displaystyle 0$ and $\displaystyle 2 \pi$. Now You use the identities...

    $\displaystyle \cos n \gamma= \frac{z^{n}+z^{-n}}{2}_{z=e^{i \gamma}}$ (2)

    $\displaystyle \cos^{2} \frac{\gamma}{2}= \frac{1+\cos \gamma}{2}$ (3)

    ... and take into Account that is $\displaystyle d \gamma=\frac{dz}{i z}$ , Your integral, if no mistakes of me, becomes...

    $\displaystyle I(x)= -\frac{i}{2 z} \int_{\Gamma} \frac{z^{n}+z^{-n}}{1+2 x^{2} + \frac{z+z^{-1}}{2}} d z = -i \int_{\Gamma} \frac{z^{2n}+1}{z^{n}\ \{1+2 (1+2 x^{2})\ z + z^{2}\}}\ dz$ (4)

    ... where $\displaystyle \Gamma$ is the unit circle. Now all that You have to do is to apply the residue theorem... difficult task but not superior to human possibilities ...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Last edited by chisigma; Sep 19th 2011 at 09:50 PM.
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  11. #11
    MHF Contributor chisigma's Avatar
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    Re: can this integral be solved analytically!

    In the previous post we are arrived to the identity...

    $\displaystyle I(x)= \int_{0}^{\pi} \frac{\cos 2 n \theta}{x^{2}+\cos^{2} \theta}\ d \theta= -i \int_{\Gamma} \frac{z^{2n}+1}{z^{n}\ \{1+2\ (1+2\ x^{2})\ z + z^{2}\}}\ dz $ (1)

    ... where $\displaystyle \Gamma$ is the unit circle. Now if we apply the residue theorem the integral (1) is given by...

    $\displaystyle I(x)= 2 \pi \sum_{k} r_{k}$ (2)

    ... where the $\displaystyle r_{k}$ are the residues of the poles of the f(z) in (1) internal to the unit circle. In $\displaystyle x \ne 0$ the f(z) has inside the unit circle a pole of order n in $\displaystyle z=0$ and a simple pole in...

    $\displaystyle z_{0}= -1 -2 x^{2} +2 x \sqrt{1+x^{2}}$ (3)

    May be is useful to proceed 'step by step', so that we start with n=0. In that case we have only the pole (3) and its residue is...

    $\displaystyle r_{1} = \lim_{z \rightarrow z_{0}} (z-z_{0})\ f(z)=\frac{1}{2 x \sqrt{1+x^{2}}} $ (4)

    ... so that is...

    $\displaystyle \int_{0}^{\pi} \frac{1}{x^{2}+\cos^{2} \theta}\ d \theta= \frac{\pi}{x\ \sqrt{1+x^{2}}}$ (5)

    In next post[s] we will consider the cases n>0...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  12. #12
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    Re: can this integral be solved analytically!

    Dear chisigma,
    Thank you very much, i just take time to check the validity of my derivation and that the integral is correct. Your solution using the residue theorem is interesting !, but what about n=1?

    Best Regards

    Mopen
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  13. #13
    MHF Contributor chisigma's Avatar
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    Re: can this integral be solved analytically!

    First we rewritre the general formula...

    $\displaystyle I(x)= \int_{0}^{\pi}\frac{\cos 2 n \theta}{x^{2}+\cos^{2} \theta}\ d \theta= -i\ \int_{\Gamma} \frac{z^{2n}+1}{z^{n}\ \{1+2 (1+2 x^{2}) z + z^{2}\}}\ dz $ (1)

    ... where $\displaystyle \Gamma$ is the unit circle. If $\displaystyle n=1$ and $\displaystyle x \ne 0$ f(z) has two poles of order 1 inside the unit circle: $\displaystyle z_{0}=0$ and $\displaystyle z_{1}=-1-2 x^{2} +2 x \sqrt{1+x^{2}}$ and their residues are...

    $\displaystyle r_{0}= \lim_{z \rightarrow z_{0}} (z-z_{0})\ f(z)= - i$ (2)

    $\displaystyle r_{1}= \lim_{z \rightarrow z_{1}} (z-z_{1})\ f(z)= -i\ \frac{1+4 x (1+x^{2}) -2 x (1+2 x^{2}) \sqrt{1+x^{2}}}{4x^{2} (1+x^{2}) -2 x (1+2 x^{2}) \sqrt{1+x^{2}}}$ (3)

    Now is...

    $\displaystyle I(x)= 2 \pi i (r_{0}+r_{1})= \pi\ \frac{1+4 x (1+x) (1+x^{2}) -4 x (1+2 x^{2}) \sqrt{1+x^{2}}}{2 x^{2} (1+x^{2}) -x (1+2 x^{2}) \sqrt{1+x^{2}}}$ (4)

    The result looks like the 'Orient Express' and , because in pure calculus I'm very poor, I suggest mopen and all the people of MHF to control my work. In any case what I can say is that for n=1 the computation efforts are remarkable and for n>1 I prefer to 'pass the ball' ...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  14. #14
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    Re: can this integral be solved analytically!

    Dear chisigma,

    May God Bless you!. you did more than enough. just small corrections for the second pole residue and the integral value " see the attached picture"

    Thank you so much!

    Best Regards

    Mopen
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