No you can't: the limit of the integral around the pole as you let e--->0 may not exist.
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 =)
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.
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 sin²(a.lnx)/(x-1)² dx=pi.a.coth(2.pi.a) - 1/2
can you help me???? members.
I am a new member from belgium
thanks
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.
Hi Era Leo,
as the integral stands its a removable singularity
but using complex contour integration yous take sin²z=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
Hi,
Era leo
concerning second order poles on the real axis and my integral from o to inf
sin²(alnx)/(x-1)²?
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)²dx+int from 1+r to inf xexp.a /(x-1)²dx -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)²
and becomes:
(1 - exp.2ia.pi)int from o to 1-r xexpa/(x-1)² +int from 1+r to inf xexp.a/(x-1)²
+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)² in a laurent serie.
f(z)=1/(x-1)²+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