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∫dx/((x^(2/3)(x+1)), integrated over [0,∞]

by Jamin2112
Tags: , ∫dx or x2 or 3x, integrated
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Jamin2112
#1
Feb5-12, 09:10 PM
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1. The problem statement, all variables and given/known data

As in thread title.

2. Relevant equations

Residue Theorem.

3. The attempt at a solution

I just need help figuring out the circle C I'll be using. Suggestions?
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vela
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Feb6-12, 04:06 AM
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What does the presence of z2/3 tell you?
Jamin2112
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Feb6-12, 09:14 AM
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Quote Quote by vela View Post
What does the presence of z2/3 tell you?
Other than that there's a pole at z=0?

vela
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Feb6-12, 10:03 AM
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∫dx/((x^(2/3)(x+1)), integrated over [0,∞]

Yes, other than that. In particular, what's the effect of the fractional power?
Jamin2112
#5
Feb6-12, 10:27 AM
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Quote Quote by vela View Post
Yes, other than that. In particular, what's the effect of the fractional power?
Change the distance between z and the origin from r to r2/3
Change the angle between z and the x-axis from to 2/3
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Feb6-12, 10:53 AM
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Right. Do you know what a branch point and a branch cut are?
Jamin2112
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Feb6-12, 12:00 PM
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Quote Quote by vela View Post
Right. Do you know what a branch point and a branch cut are?
Yeah, I somehow need a loop that avoid z=-1 and z=0. Right?
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Feb6-12, 12:03 PM
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It's more that you want to avoid crossing the branch cut than avoiding z=0, and you obviously want a piece or pieces of the contour to correspond to the original integral.
Jamin2112
#9
Feb6-12, 12:22 PM
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Quote Quote by vela View Post
It's more that you want to avoid crossing the branch cut than avoiding z=0, and you obviously want a piece or pieces of the contour to correspond to the original integral.
So I'd take R>1 and make a half circle of radius R in the upper half of the plane. Then I'd make two little half circles that jump over z=-1 and z=0. Then I'd look at ∫C f(z)dz as the sum of several integrals, one of which can written as a real-valued integral and see what happens as R→∞ and the radii of the little half circles go to zero. Right?
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Feb6-12, 12:36 PM
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No, that's too complicated. Take a look at http://en.wikipedia.org/wiki/Methods...93_branch_cuts.

Also, rewrite the integrand as
$$\frac{z^{1/3}}{z(z+1)}$$to make it clear how to calculate the residue at z=0.
Jamin2112
#11
Feb6-12, 12:42 PM
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Quote Quote by vela View Post
No, that's too complicated. Take a look at http://en.wikipedia.org/wiki/Methods...93_branch_cuts.

Also, rewrite the integrand as
$$\frac{z^{1/3}}{z(z+1)}$$to make it clear how to calculate the residue at z=0.
So in my case the path should resemble a backwards Pacman?
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Feb6-12, 02:16 PM
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Doesn't the answer to that question depend on which way Pacman is moving?
Jamin2112
#13
Feb6-12, 05:21 PM
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Doesn't the answer to that question depend on which way Pacman is moving?
I forgot that PacMan is in perpetual motion.

But yeah, how am I gonna do this? I need C to be formed from a series of paths, each of which will have a line integral that approaches a real value after I take some limit.
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#14
Feb6-12, 06:04 PM
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Start by taking the keyhole contour and break it into four pieces and evaluate the line integral for each piece.
Jamin2112
#15
Feb6-12, 06:24 PM
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Start by taking the keyhole contour and break it into four pieces and evaluate the line integral for each piece.
How would that work? I want ∫f(x)dx (integrated on [0, R]) to be one of the four line integrals.
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Feb8-12, 02:47 PM
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That's what you're supposed to figure out. Did you understand the example on Wikipedia? That's pretty much the recipe you want to follow.


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