Results 1 to 14 of 14

Math Help - can this integral be solved analytically!

  1. #1
    Newbie
    Joined
    Sep 2011
    Posts
    24

    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
    Attached Thumbnails Attached Thumbnails can this integral be solved analytically!-integral.png  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    3

    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,

    \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
    .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2011
    Posts
    24

    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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: can this integral be solved analytically!

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Sep 2011
    Posts
    24

    Re: can this integral be solved analytically!

    But what does it means, Seriesexpansion of the integral at :t =‡ 0?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5

    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...

    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...

    \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

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Sep 2011
    Posts
    24

    Re: can this integral be solved analytically!

    i prefer "a" , but how to substitute the cos(2nθ) term?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5

    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...

    \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

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Sep 2011
    Posts
    24

    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?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5

    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 \gamma=2 \theta so that the integral becomes...

    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 0 and 2 \pi. Now You use the identities...

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

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

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

    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 \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

    \chi \sigma
    Last edited by chisigma; September 19th 2011 at 09:50 PM.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5

    Re: can this integral be solved analytically!

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

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

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

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

    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...

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

    ... so that is...

     \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

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Sep 2011
    Posts
    24

    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
    Follow Math Help Forum on Facebook and Google+

  13. #13
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    5

    Re: can this integral be solved analytically!

    First we rewritre the general formula...

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

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

    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...

    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

    \chi \sigma
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Newbie
    Joined
    Sep 2011
    Posts
    24

    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
    Attached Thumbnails Attached Thumbnails can this integral be solved analytically!-new-picture.png  
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: June 4th 2011, 12:11 PM
  2. Analytically find the value of sin(x)
    Posted in the Calculus Forum
    Replies: 13
    Last Post: May 19th 2011, 09:38 PM
  3. Analytically find the value of sin(x)
    Posted in the Trigonometry Forum
    Replies: 6
    Last Post: May 19th 2011, 08:45 AM
  4. Solve Integral Analytically
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 21st 2009, 11:50 AM
  5. Limits, Analytically
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 11th 2008, 03:30 PM

Search Tags


/mathhelpforum @mathhelpforum