Question

Complex analysis

Answer 1

vague memory...

Answer 2

\[\int\limits_{0}^{\infty} x ^{p}/(x ^{2}+1)(x+1)^{2} dx \]

Answer 3

-1<p<3

Answer 4

residues? damn. integral is 2 pi i times sum of residues? i should be quiet or look in a book

Answer 5

it is, but I don't know how to find residues.

Answer 6

if i remember correctly for a rational function it is
\[\frac{f(z_0)}{g'(z_0)}\]

Answer 7

do u know how to solve a problem above?

Answer 8

i am going to attempt it and will shut up until i think i have an answer.

Answer 9

ok :D thanks :)

Answer 10

You'll find a worked example here in this article.
http://en.wikipedia.org/wiki/Methods_of_contour_integration#Example

See if this makes sense and try and imitate it for your problem

Answer 11

i found the residue at x = 1 is in fact 0

Answer 12

actually yours is example IV the "keyhole" integral

Answer 13

James, this is some other type of integral :(

Answer 14

look at example IV

Answer 15

yes, that's it

Answer 16

why 1 is residue?

Answer 17

because i made a typo. -1 is a pole, not 1

Answer 18

so, do u know how to solve that?

Answer 19

actually no, because i have no idea what to do with that
\[x^p\] in the numerator. i guess your answer will have a p in it, but i don't know what or where

Answer 20

simple pole at i and -i, and pole of order two at -1
residue at i is
\[\frac{i^p}{4i^3+6i^2+4i+2}=\frac{i^p}{-4}\]

Answer 21

I'm running around this morning and need to step out for a bit. I can get back to this later and I'd like to, because it's a fun problem.

Answer 22

get busy. have work to do as well, but at i did get one residue.
@gg the denominator is the derivative of your original denominator evaluated at i. i am going to bet we get the same thing evaluated at - i

Answer 23

James, I would be soooooo thankful, I have exam tomorrow and I don't know how to do this. Could u be back in 1-2 hours?

Answer 24

I'm going to look in a book now, I'll see what I can do... if I can do anything :/

Answer 25

yes it is the same. residue at -i is also
\[\frac{i^p}{-4}\]

Answer 26

ok, but how to find residues? Can u show me step by step?

Answer 27

yes that much i can do

Answer 28

for
\[\frac{f(x)}{g(x)}\] the residue at a pole z0 (zero of g) is
\[\frac{f(z_0)}{g'(z_0)}\]

Answer 29

Just so I'm clear, are we talking about:
\[\int_{0}^{\infty}\frac{x^p}{(x^2+1)(x+1)^2}dx\]

Answer 30

that is if it is a zero of multiplicity 1

Answer 31

yes that is the integral.

Answer 32

what is f and what is g?

Answer 33

in this case you have
\[\frac{x^p}{(x^2+1)(x+1)^2}\] so
\[f(x)=x^p, g(x)=(x^2+1)(x+1)^2\]

Answer 34

i just mean numerator and denominator.

Answer 35

in every example f is numerator and g is denominator?

Answer 36

valpey, u were writing something?

Answer 37

yes, you find the zero of the denominator, evaluate at the numerator divided by the derivative of the denominator. but that is for a simple pole, a zero of multiplicity 1

Answer 38

I'm making notes. I don't have anything for you yet.

Answer 39

now just have to remember how to get the residue of a zero of multiplicity 2

Answer 40

I go to try to do this, I'll be back later. If u have something for me, write please :) I don't wanna fail exam AGAIN :D

Answer 41

Okay, so singularities at
\[\pm i, -1\]
Maybe the branch cut contour they use in example iv of the Methods of Contour wiki.
... well, I pretty much failed most of my complex analysis exams, but maybe this output from Wolfram Alpha will ring a bell for you:
\[\int_{0}^{\infty} \frac{x^p}{(1+x)^2 (1+x^2)} dx = \frac{1}{4} \pi (csc(\frac{\pi p}{2})+2 (p-1) csc(\pi p)) ::for -1<Re(p)<3\]
Good luck!

Answer 42

what is csc?

Answer 43

cosecant = 1/sin(x)

Answer 44

it still doesn't help... I need to know HOW to solve this.

Answer 45

Yeah. I'm with you. In terms of the contours though, I think you need a counter-clockwise circle with radius > 1 and a clockwise circle with radius < 1
\[\int_{0}^{\infty} \frac{x^p}{(1+x)^2 (1+x^2)} dx =\int_{C} \frac{z^p}{(1+z)^2 (1+z^2)} dz \]

\[=\int_{C_1}^{\infty} \frac{z^p}{(1+z)^2 (1+z^2)} dz +\int_{C_2}^{\infty} \frac{z^p}{(1+z)^2 (1+z^2)} dz\]