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Math Help - contour integral, limiting contour theorem with residue

  1. #1
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    contour integral, limiting contour theorem with residue

    Hi, can anyone please help me with this question, thanks a lot.

    \displaystyle \int_{0}^{\infty} \frac{x^{-1/3}}{x^2+1} dx

    I did try to take the contour, and take notice the three "bad points" are 0, i, and -i.

    I used residue theorem that \displaystyle\oint_{\Gamma_{R,\epsilon}} \frac{dz}{\sqrt[3]{z}(z^2+1)}=2\pi i\displaystyle \sum_{poles\ in\ the\ plane}Res(f(z), a_j).

    I can use limiting contour theorem to get one integral is 0.

    However, I'm really having trouble solve this question, I thought my methods are right, but I can't get the right answer which is \frac{\sqrt{3}{\pi}}{3}. One friend told me I need to worry about choosing branch because of that \sqrt[3]{z}, but I don't quite understand it and what I supposed to do.

    Can anyone please show me some precise steps on solving this question? Thanks a lot.
    Last edited by tsang; May 21st 2011 at 12:19 AM. Reason: typo
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  2. #2
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by tsang View Post
    Hi, can anyone please help me with this question, thanks a lot.

    \displaystyle \int_{0}^{\infty} \frac{x^{-1/3}}{x^2+1} dx

    I did try to take the contour, and take notice the three "bad points" are 0, i, and -i.

    I used residue theorem that \displaystyle\oint_{\Gamma_{R,\epsilon}} \frac{dz}{\sqrt[3]{z}(z^2+1)}=2\pi i\displaystyle \sum_{poles\ in\ the\ plane}Res(f(z), a_j).

    I can use limiting contour theorem to get one integral is 0.

    However, I'm really having trouble solve this question, I thought my methods are right, but I can't get the right answer which is \frac{\sqrt{\pi}}{3}. One friend told me I need to worry about choosing branch because of that \sqrt[3]{z}, but I don't quite understand it and what I supposed to do.

    Can anyone please show me some precise steps on solving this question? Thanks a lot.
    The function to be integrated has two 'simple poles' in z=i and z=-i and one 'brantch point' in z=0. The last type of singularity has to be excluded from the integration path so that may be that the best integration path is illustrated in the figure...



    Kind regards

    \chi \sigma
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  3. #3
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    Quote Originally Posted by chisigma View Post
    The function to be integrated has two 'simple poles' in z=i and z=-i and one 'brantch point' in z=0. The last type of singularity has to be excluded from the integration path so that may be that the best integration path is illustrated in the figure...



    Kind regards

    \chi \sigma

    Hi, thank you for your help. Yes, I used the same contour as your graph.
    But, could you please give me a bit more details? I'm still confused on what I should do. I can get the whole contour is made by four parts, but I still can't get the right answer. Please help me a bit more. Thanks a lot.
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by tsang View Post
    Hi, thank you for your help. Yes, I used the same contour as your graph.
    But, could you please give me a bit more details? I'm still confused on what I should do. I can get the whole contour is made by four parts, but I still can't get the right answer. Please help me a bit more. Thanks a lot.
    All right!... may be it is useful to examine the more general case...

    I= \int_{0}^{\infty} \frac{x^{p}}{1+x^{2}}\ dx\ ,\ -1<p<0 (1)

    As said above the solution is in the computation of the integral...

    \int_{\gamma} f(z)\ dz = \int_{\gamma} \frac{z^{p}}{1+z^{2}}\ dz (2)

    ... along the 'red path' \gamma of the figure...



    The procedure is a lottle long and it is better to devide it into two steps. The first step is the computation of the integral (2) that is done by the Cauchy integral formula...

    \int_{\gamma} f(z)\ dz = 2 \pi\ i\ \sum_{i} r_{i} (3)

    In our case f(z) inside \gamma has two simple poles: z=i and \z=-i and their residues are...

    r_{1}= \lim_{z \rightarrow i} (z-i)\ \frac{z^{p}}{1+z^{2}}= \frac{e^{i\ \frac{\pi\ p}{2}}}{2i} (4)

    r_{2}= \lim_{z \rightarrow -i} (z+i)\ \frac{z^{p}}{1+z^{2}}= -\frac{e^{-i\ \frac{\pi\ p}{2}}}{2i} (5)

    ... so that the integral (3) is...

    \int_{\gamma} f(z)\ dz = 2 \pi\ i\ \sum_{i} r_{i}= 2 \pi\ i\ \sin \frac{\pi\ p}{2} (6)

    ... and the first step is done. Are You able to proceed?...

    Kind regards

    \chi \sigma
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by chisigma View Post
    All right!... may be it is useful to examine the more general case...

    I= \int_{0}^{\infty} \frac{x^{p}}{1+x^{2}}\ dx\ ,\ -1<p<0 (1)

    As said above the solution is in the computation of the integral...

    \int_{\gamma} f(z)\ dz = \int_{\gamma} \frac{z^{p}}{1+z^{2}}\ dz (2)

    ... along the 'red path' \gamma of the figure...



    The procedure is a lottle long and it is better to devide it into two steps. The first step is the computation of the integral (2) that is done by the Cauchy integral formula...

    \int_{\gamma} f(z)\ dz = 2 \pi\ i\ \sum_{i} r_{i} (3)

    In our case f(z) inside \gamma has two simple poles: z=i and \z=-i and their residues are...

    r_{1}= \lim_{z \rightarrow i} (z-i)\ \frac{z^{p}}{1+z^{2}}= \frac{e^{i\ \frac{\pi\ p}{2}}}{2i} (4)

    r_{2}= \lim_{z \rightarrow -i} (z+i)\ \frac{z^{p}}{1+z^{2}}= -\frac{e^{-i\ \frac{\pi\ p}{2}}}{2i} (5)

    ... so that the integral (3) is...

    \int_{\gamma} f(z)\ dz = 2 \pi\ i\ \sum_{i} r_{i}= 2 \pi\ i\ \sin \frac{\pi\ p}{2} (6)

    ... and the first step is done...
    The second step is the division of integral (6) into four distinct integrals...

    \int_{\gamma} f(z)\ dz = \int_{r}^{R} \frac{x^{p}}{1+x^{2}}\ dx + i\ \int_{0}^{2 \pi} \frac{R^{p+1}\ \theta\ e^{i\ \theta\ (p+1)}}{1+R^{2}\ e^{2 i \theta}}\ d \theta +

    + \int_{R}^{r} \frac{x^{p}\ e^{2\ \pi\ i\ p}}{1+x^{2}\ e^{4\ \pi\ i}}\ dx + i\ \int_{2\ \pi}^{0} \frac{r^{p+1}\ \theta\ e^{i\ \theta\ (p+1)}}{1+r^{2}\ e^{2 i \theta}}\ d \theta (7)

    Now if -1<p<0 the second integral in (7) vanishes if R tends to infinity and the fourth integral in (7) also vanishes if r tends to 0, so that, taking into account (6), is...

    (1- e^{2\ \pi\ i\ p})\ \int_{0}^{\infty} \frac{x^{p}}{1+x^{2}}\ dx = 2\ \pi\ i\ \sin \frac{\pi\ p}{2} (8)

    ... and from (8) with simple steps we arrive at the result...

    \int_{0}^{\infty} \frac{x^{p}}{1+x^{2}}\ dx = (-1)^{1-p}\ \pi\ \frac{\sin \frac{\pi\ p}{2}}{\sin \pi\ p} (9)

    The result (9) is in some way 'a little enbarassing'... for p=-\frac{1}{3} we have the correct result...

    \int_{0}^{\infty} \frac{x^{-\frac{1}{3}}}{1+x^{2}}\ dx = \frac{\pi}{\sqrt{3}} (10)

    ... as in...

    int x&#94;&#40;-1&#47;3&#41;&#47;&#40;1&#43;x&#94;2&#41; dx, x&#61;0..infinity - Wolfram|Alpha

    ... but for other values of p (9) [for istance p=-\frac{1}{2} ...] that is not true... an interesting problem for the 'experts' [unless some mistake of me]...

    Kind regards

    \chi \sigma
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  6. #6
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    Quote Originally Posted by chisigma View Post
    The second step is the division of integral (6) into four distinct integrals...

    \int_{\gamma} f(z)\ dz = \int_{r}^{R} \frac{x^{p}}{1+x^{2}}\ dx + i\ \int_{0}^{2 \pi} \frac{R^{p+1}\ \theta\ e^{i\ \theta\ (p+1)}}{1+R^{2}\ e^{2 i \theta}}\ d \theta +

    + \int_{R}^{r} \frac{x^{p}\ e^{2\ \pi\ i\ p}}{1+x^{2}\ e^{4\ \pi\ i}}\ dx + i\ \int_{2\ \pi}^{0} \frac{r^{p+1}\ \theta\ e^{i\ \theta\ (p+1)}}{1+r^{2}\ e^{2 i \theta}}\ d \theta (7)

    Now if -1<p<0 the second integral in (7) vanishes if R tends to infinity and the fourth integral in (7) also vanishes if r tends to 0, so that, taking into account (6), is...

    (1- e^{2\ \pi\ i\ p})\ \int_{0}^{\infty} \frac{x^{p}}{1+x^{2}}\ dx = 2\ \pi\ i\ \sin \frac{\pi\ p}{2} (8)

    ... and from (8) with simple steps we arrive at the result...

    \int_{0}^{\infty} \frac{x^{p}}{1+x^{2}}\ dx = (-1)^{1-p}\ \pi\ \frac{\sin \frac{\pi\ p}{2}}{\sin \pi\ p} (9)

    The result (9) is in some way 'a little enbarassing'... for p=-\frac{1}{3} we have the correct result...

    \int_{0}^{\infty} \frac{x^{-\frac{1}{3}}}{1+x^{2}}\ dx = \frac{\pi}{\sqrt{3}} (10)

    ... as in...

    int x&#94;&#40;-1&#47;3&#41;&#47;&#40;1&#43;x&#94;2&#41; dx, x&#61;0..infinity - Wolfram|Alpha

    ... but for other values of p (9) [for istance p=-\frac{1}{2} ...] that is not true... an interesting problem for the 'experts' [unless some mistake of me]...

    Kind regards

    \chi \sigma
    Hi, sorry I just realised there's something confusing from your (8) to (10). This is because I do understand everything you did up to part (8), but I can't see how you get from part (8) to part (9), then, I used Matlab to double check, if I subsititute p=-\frac{1}{3}in your equation (9), I don't actually get \frac{\pi}{\sqrt{3}}, it is different answer.
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  7. #7
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by tsang View Post
    Hi, sorry I just realised there's something confusing from your (8) to (10). This is because I do understand everything you did up to part (8), but I can't see how you get from part (8) to part (9), then, I used Matlab to double check, if I subsititute p=-\frac{1}{3}in your equation (9), I don't actually get \frac{\pi}{\sqrt{3}}, it is different answer.
    In my opinion for -1<p < 1 is 'with great probability'...

    \int_{0}^{\infty} \frac{x^{p}}{1+x^{2}}\ dx = \pi\ \frac{\sin \frac{\pi\ p}{2}}{\sin \pi\ p} = \frac{\frac{\pi}{2}}{\cos \frac{\pi\ p}{2}} (1)

    For example p=\pm \frac{1}{2} gives...

    \int_{0}^{\infty} \frac{x^{\pm \frac{1}{2}}}{1+x^{2}} = \frac{\pi}{\sqrt{2}} (2)

    ... as in...

    int x&#94;&#40;1&#47;2&#41;&#47;&#40;1&#43;x&#94;2&#41 ; dx, x&#61;0..infinity - Wolfram|Alpha

    ... and in...

    int x&#94;&#40;-1&#47;2&#41;&#47;&#40;1&#43;x&#94;2&#41; dx, x&#61;0..infinity - Wolfram|Alpha

    My only problem is... I am unable to demonstrate that ... if the result I arrived to is correct...

     (1-e^{2 \pi i p})\ \int_{0}^{\infty} \frac{x^{p}}{1+x^{2}}\ dx = 2 \pi i \sin \frac{\pi p}{2} (3)

    ... then I'm in a trouble because is...

    1-e^{2 \pi i p} = - e^{\pi i p}\ (e^{\pi i p}-e^{-\pi i p}) = 2 i e^{\pi i (p-1)}\ \sin \pi p (4)

    ... and the fastidious term e^{\pi i (p-1)}= (-1)^{p-1} is not eliminable... that's why I'm expecting some 'help' from the 'experts' of the forum...

    Kind regards

    \chi \sigma
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