Results 1 to 11 of 11

Math Help - improper integral using complex analysis

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    5

    improper integral using complex analysis

    hi everyone, i have a question here regarding to my topic.

    I know that we could solve improper integral from -infinity to infinity by drawing a contour around the real axis and a big semi circle with R-->infinity to join the end of real infinity and real -infinity. however if there is a pole on the real axis, we need to make a small semi circle to 'skip' the pole.

    The integration of this small semi circle on a single pole is given by (pi)(i)(Res(f,a)) where a is the position of the pole in the real axis.

    May i know how should i approach the similar question but with repeated pole on point a now? can i still using the formula (pi)(i)(Res(f,a)) ?

    I have checked the prove for this formula, it is initially assume that there is only a single pole at point a. That's why I not sure whether I could apply the same formula for a repeated pole?

    Thank you everyone =)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Oli
    Oli is offline
    Member
    Joined
    Apr 2008
    Posts
    82
    No you can't: the limit of the integral around the pole as you let e--->0 may not exist.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2008
    Posts
    5
    so does that means that there are no way to solve an improper integral with a 2nd order real pole?

    thanks alot =)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Oli
    Oli is offline
    Member
    Joined
    Apr 2008
    Posts
    82
    Well, if you are integrating from -inf to +inf that is a proper integral.

    I think it can be done in some cases with pole of order 2+, but it just needs special consideration.

    Since you get order z^-n integrals to do, where n>1, I would guess you have to look at each of these terms and do something clever. But these terms are going to get big pretty quickly as z goes to 0, so it is very probable these terms won't converge. In fact I would say that it is almost certain they won't converge.

    I haven't really done much of this for ages, but you can probably see the problem that arises.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    You can also try doing some substitution trick which will change the integral into the one which has a simple pole at the origin. Do you have any example?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Apr 2008
    Posts
    5
    I don't have any examples yet. Actually in my current course I will only deal with simple pole on real axis. But my Prof challenge me to derive a general formula for a repeated pole on real axis. That's how this question is here.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Apr 2008
    Posts
    5
    does anyone has an example of second order pole function which is convergent? From my derivative, all the function with second or higher order pole are not convergent. Is that true?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Apr 2008
    Posts
    24

    second order pole on real axis

    hello Era Elo,

    Maybe you can try partial integration an see if you become a limit.Then the pole would be in some circumstances of the first order.

    I have the same problem with the integral:

    int from 0 to infinity sin(a.lnx)/(x-1) dx=pi.a.coth(2.pi.a) - 1/2

    can you help me???? members.

    I am a new member from belgium

    thanks
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Apr 2008
    Posts
    5
    hey gilbert

    i haven tried to attempt ur question here but i have noticed something.
    this function has NO second order real pole. because the pole is canceled by the sine function at the numerator. Hence it is actually a removable pole which means, no pole at all. mayb u will find this useful.

    if u want to check what is the order of the pole, there is this method here:

    substitute the pole position(in this case, 1) into the numerator, and see if the numerator goes to zero. if it goes to zero, then take the first derivative of the numerator and then substitute 1 again into the equation. if u get zero again, differentiate it again and substitute again. until u get something which is NOT zero. then the number of zero u got is the order of the roots(zero).

    then is obvious in the denominator u have 2nd order pole at position 1, so the overall order of pole is the order of zero minus order of pole. if u get anything 0 or below, means ur function has removable pole.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Apr 2008
    Posts
    24

    inmproper integrals using complex analysis

    Hi Era Leo,

    as the integral stands its a removable singularity

    but using complex contour integration yous take sinz=1/2(1-cos(2z)

    or (exp 2ialnz - 2 + exp-2ialnz)/-4 and with - 2 oy have a second order pole
    I tried with the keyhole contour without result
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Apr 2008
    Posts
    24

    improper intergral using complex analysis

    Hi,
    Era leo

    concerning second order poles on the real axis and my integral from o to inf

    sin(alnx)/(x-1)?
    In a bib I consult a book on complex variables by F Bayen and C Margaria.They
    give a solution for a second order pole on the real axis.

    lim r->0{int from 0 to 1-r x.expa/(x-1)dx+int from 1+r to inf xexp.a /(x-1)dx -2/r}= -pi.a.cotg pi.a (I)

    We use the keyhole contour with the branch 0<phi<2.pi
    and with an identitation on the upper cut an lower cut.

    W integrate around the keyhole contour the function f(z)=z.expa/(x-1)

    and becomes:

    (1 - exp.2ia.pi)int from o to 1-r xexpa/(x-1) +int from 1+r to inf xexp.a/(x-1)

    +int from pi to 0 f(1+r exp.i.phi)irexp i.phi d phi+int from 0 to - pi f(1+rexp.iphi)irexp i.phi=0

    by cauchys theorem.
    r=radius of half circles at 1
    phi=argument z

    i

    Now we expand z.expa/(z-1) in a laurent serie.

    f(z)=1/(x-1)+a/(x-1)+g(z) (analytic) function

    integrate on the upper half circle

    we have 2/r + pi.i.a +0 (cauchy theorem-analytic function g(z)

    integrate on the lower cut

    we have exp 2.i.a.pi (-2/r+i.a.pi+o)

    In the limit r->0 we get the standard expression (I) above

    Now the problem is the ter 2/r in (I)

    Now problem for the calculation of an integral as in my example they insert

    a convergence factor:let us take without loss of generality x.expb and then
    in teresult we take the lim b->0.

    for my example -1/4(xexp 2ia -2 +xexp-2ia)

    remember cotg(ix)=-i coth(x)

    cotg(-x)=-cotg(x)

    We use now expression (I) with a=2ia and a=-2ia

    then we have:

    -1/'4pi{-2ia+b)cotg(2iab)+2bcotg b-(-2ia+b)cotg(-2ia+b)}in this expr. you have -2/r+4/r-2/r see formule (I) the singular terms cancel in the limit

    take lim b->0

    =-1/4 pi{-4acoth(2pi.a) +2}

    =pi.a coth (2.pi.a) - 1/2 what the desired result is.

    Ik know its a bit difficult to write the mathematical expressions.

    How can I type math expressions:software or math keyboard.

    I think the method can be used for higher order poles on the real axis.

    Maybe your prof. have an other method to calculate such integrals??????

    If you will can you tell me what he says of the method here.

    thanks
    Last edited by Gilbert; April 15th 2008 at 04:09 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Complex Integral in Complex Analysis
    Posted in the Advanced Math Topics Forum
    Replies: 1
    Last Post: August 21st 2011, 09:46 PM
  2. Real integral with complex analysis
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 23rd 2010, 01:32 PM
  3. [SOLVED] Calculating an improper integral via complex analysis
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: October 25th 2009, 11:33 AM
  4. complex analysis integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 6th 2007, 12:44 PM
  5. Complex Analysis-- Integral
    Posted in the Calculus Forum
    Replies: 10
    Last Post: December 21st 2006, 09:02 AM

Search Tags


/mathhelpforum @mathhelpforum