Results 1 to 7 of 7

Thread: contour integral, limiting contour theorem with residue

  1. #1
    Member
    Joined
    Jul 2010
    Posts
    99

    contour integral, limiting contour theorem with residue

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

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

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

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

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

    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 $\displaystyle \frac{\sqrt{3}{\pi}}{3}$. One friend told me I need to worry about choosing branch because of that $\displaystyle \sqrt[3]{z}$, 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 20th 2011 at 11:19 PM. Reason: typo
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    6
    Quote Originally Posted by tsang View Post
    Hi, can anyone please help me with this question, thanks a lot.

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

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

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

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

    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 $\displaystyle \frac{\sqrt{\pi}}{3}$. One friend told me I need to worry about choosing branch because of that $\displaystyle \sqrt[3]{z}$, 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

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

  3. #3
    Member
    Joined
    Jul 2010
    Posts
    99
    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

    $\displaystyle \chi$ $\displaystyle \sigma$

    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    6
    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...

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

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

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

    ... along the 'red path' $\displaystyle \gamma$ 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...

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

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

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

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

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

    $\displaystyle \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

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

  5. #5
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    6
    Quote Originally Posted by chisigma View Post
    All right!... may be it is useful to examine the more general case...

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

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

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

    ... along the 'red path' $\displaystyle \gamma$ 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...

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

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

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

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

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

    $\displaystyle \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...

    $\displaystyle \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 + $

    $\displaystyle + \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 $\displaystyle -1<p<0$ 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...

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

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

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

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

    $\displaystyle \int_{0}^{\infty} \frac{x^{-\frac{1}{3}}}{1+x^{2}}\ dx = \frac{\pi}{\sqrt{3}}$ (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 $\displaystyle p=-\frac{1}{2}$ ...] that is not true... an interesting problem for the 'experts' [unless some mistake of me]...

    Kind regards

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

  6. #6
    Member
    Joined
    Jul 2010
    Posts
    99
    Quote Originally Posted by chisigma View Post
    The second step is the division of integral (6) into four distinct integrals...

    $\displaystyle \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 + $

    $\displaystyle + \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 $\displaystyle -1<p<0$ 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...

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

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

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

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

    $\displaystyle \int_{0}^{\infty} \frac{x^{-\frac{1}{3}}}{1+x^{2}}\ dx = \frac{\pi}{\sqrt{3}}$ (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 $\displaystyle p=-\frac{1}{2}$ ...] that is not true... an interesting problem for the 'experts' [unless some mistake of me]...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    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 $\displaystyle p=-\frac{1}{3}$in your equation (9), I don't actually get $\displaystyle \frac{\pi}{\sqrt{3}}$, it is different answer.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    6
    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 $\displaystyle p=-\frac{1}{3}$in your equation (9), I don't actually get $\displaystyle \frac{\pi}{\sqrt{3}}$, it is different answer.
    In my opinion for $\displaystyle -1<p < 1$ is 'with great probability'...

    $\displaystyle \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 $\displaystyle p=\pm \frac{1}{2} $ gives...

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

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

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

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

    Kind regards

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

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Aug 31st 2010, 07:38 AM
  2. Integral over head-light contour via Residue Theorem
    Posted in the Math Challenge Problems Forum
    Replies: 0
    Last Post: Feb 5th 2010, 02:39 PM
  3. contour integral and residue thm.
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: Aug 27th 2009, 05:33 AM
  4. Replies: 0
    Last Post: May 5th 2009, 12:30 PM
  5. Replies: 2
    Last Post: Jan 17th 2009, 03:51 AM

Search Tags


/mathhelpforum @mathhelpforum