Evaluating Higher Order Poles: A Simple Download

[Living_Dog]

I need to know how to evaluate higher order poles.

I have the answer for the integral of this function

$$\frac{(1 + x^2)}{(1 + x^4)}[\tex]<br /> <br /> from integrals.wolfram.com, but think it can be done using residues. I believe it involves taking a derivative and then multiplying by the pole? Can someone give me the simple download on what the math technique is... without all the high-dimensional mathematical proof... if you don't mind. <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f615.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":confused:" title="Confused :confused:" data-smilie="5"data-shortname=":confused:" /><br /> <br /> <i>Thankis in advance!</i><br /> -LD<br /> ________________________________________________<br /> http://www.angelfire.com/ny5/jbc33/"$$

 

[gammamcc]

Wait.. What is the region of integration..
Your tex did not come out...

 

[HallsofIvy]

End tex with [/ tex], not [\tex] (and remove the space).

Living Dog, I assume this is over a closed path but it is crucial that we know what poles are contained inside the path.

I don't see why you are asking about "higher order" poles. x⁴+ 1 factors as (x+1)(x-1)(x+i)(x-i) and has each of 1, -1, i, and -i as a simple pole. Saying that a function has a simple pole at x= a means that it can be written F(x)= f(x)+ A/(x-a) where f(x) is analytic at x= a. The integral of an analytic function around any close curve is 0 so the integral of F(x) around any closed curve containing a (and no other poles) is equal to the integral of A/(x-a). Of course, that would be the same for any closed path containing a. In particular, we can take a circular path with center a and radius R. On that curve, $x= a+ Re^{i\theta}$ so $(x- a)^{-1}= R^{-1}e^{-i\theta}$ and $dx= iRe^{i\theta}d\theta$. That is:
$$\int A/(x-a)dx= i\int_0^{2\pi}d\theta= 2\pi iA$$
Since F(x)= f(x)+ A/(x-a), F(x)(x-a)= f(x)(x-a)+ A. f(x) is analytic so its limit as x-> a is 0 and we have A= lim(x->a)F(x)(x-a).

If F does have a higher order pole, say of order n, then it can be written as a "Laurent" series- a series involving fractional powers to power n:
$$F(x)= f(x)+ A/(x-a)+ B/(x-a)^2+\cdot\cdot\cdot+ Z/(x-a)^n$$
But it is still true that the integral, around any closed path containing a as its only pole, of every term except A(x-a) is 0 and the integral of A/(x-a) is $2\pi i A$. To find A multiply by (x-a)ⁿ:
$$F(x)(x-a)^n= f(x)(x-a)^n+ A(x-a)^{n-1}+ \cdot\cdot\cdot+ Z$$
Now differentiate n-1 times:
$$\frac{d^{n-1}[F(x)(x-a)^n]}{dx^{n-1}}= \frac{d^{n-1}[f(x)(x-a)^n]}{dx^{n-1}}+ (n-1)!A$$
Since f(x) is analytic, so is f(x)(x-a)^n-1 and so its' n-1 derivative is continuous: its limit at a is 0. We have
$$A= \frac{1}{(n-1)!} \lim_{x\rightarrow a}\frac{d^{n-1}[F(x)(x-a)^{n-1}]}{dx^{n-1}}$$
for the residue at a pole of order n.

 

[mathwonk]

these are simple poles. i.e. since 1+x^4 has simple zeroes, dividing by it produces simple poles.