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Hi Scott, I am learning complex analysis on my own and am working through a section on complex integration. I have a good grasp on the section, but I have a question about a type of problem that wasn't in the problem set. They have integrals such as the integral of x/(x^3+1) dx from 0 to infinity. The examples in the book describe a method for using the substitution x=e^t. I understand the method, but the book doesn't have anything about poles with order > 1. For instance the integral of x/(x^3+1)^2 dx from 0 to infinity. How would I handle something like this?

Thanks!

Randy here. Choosing the contour for this integral is a bit tricky although dealing with the fact that the pole is 2nd order isn't necessarily hard (i'll get to that in a minute). Using the substitution x = e^t may work, but it moves the poles around and makes things complicated; I haven't pursued it.

One trick I came across is to draw the contour in 3 pieces:

1) from 0 -> infinity along the x-axis, just like the limits of the integral; call the integral for this segment I1
2) I2 = integral along the usual arc (that will extend out to infinity) from +x counter clockwise to the line given by...
3) xe^(2𝛑/3) as x goes from infinity to 0.

This last piece, which is a line heading off at an angle of 120 degrees, serves 2 purposes:
a) its closes the contour so it contains only the one pole at i𝛑/3 (so there is only one residue to evaluate)
b) the integral along it can be written as a constant times the integral of interest (i.e., along the positive x axis),

segment 3) integrand = { [xe^(i2𝛑/3)][e^(i2𝛑/3)dx] ]/[(xe^i2𝛑/3)^3+1]^2 = -e^(i4𝛑/3)I1

where we have used [e^(i2𝛑/3)]^3 = 1 in the denominator and the minus sign takes into account the opposing integration directions.

So, the contour integral is 2𝛑iRes(z=e^(i𝛑/3)) = I1 - e^(i4𝛑/3)I1 + I2.

Now, since the denominator is of a sufficiently high power of x compared to the numerator, I2 -> 0 as the radius of the arc -> infinity.

So we "just" need to find Res(z). The usual textbook approach should work here where the first derivative of (z-e^(i𝛑/3)^2･integrand is evaluated at z=e^(i𝛑/3).

To accomplish this, the denominator is expanded in terms of its roots, (z^3+1)^2 = (z-e^i𝛑/3)^2･(z-e^(i𝛑))^2･(z-e^(i5𝛑/3)･^2; the (z-e^(i･/3)^2 term cancels out but now you still need to calculate the derivative (including an x in the numerator).

Good luck!

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 Rating(1-10) Knowledgeability = 10 Clarity of Response = 10 Politeness = 10 Comment Thanks,Randy! I appreciate it.

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#### randy patton

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college mathematics, applied math, advanced calculus, complex analysis, linear and abstract algebra, probability theory, signal processing, undergraduate physics, physical oceanography

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26 years as a professional scientist conducting academic quality research on mostly classified projects involving math/physics modeling and simulation, data analysis and signal processing, instrument development; often ocean related

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J. Physical Oceanography, 1984 "A Numerical Model for Low-Frequency Equatorial Dynamics", with M. Cane

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