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ReEvaluating real integrals using residue calculus

  1. Aug 31, 2006 #1

    I recently came to know that some real integrals become easier to evaluate using the techniques of residue calculus from complex analysis. I thought that this would be a good tool to pickup apart from the standard techniques of by parts and substitution that we are taught. I have a few questions in this regard.

    1)How can I recognise an integral that will be more easily evaluated using residue calculus than ordinary methods?
    2)How do I find out if the complex arc part of the contour integral vanishes?
    3)Can you give me a few simple problems to practice my rudimentary knowledge?


  2. jcsd
  3. Aug 31, 2006 #2
    1) There are some standard formats/ combinations of functions.
    2) Sorry, I am not very clear about this question.
    3) Question: [tex] I = \int_{0}^{\Pi} \frac{1}{{(a+b \cosx)^2}} dx
    Answer : [tex]I = \frac {ab^2 \Pi}{(\sqrt{a^2-b^2})^3}[/tex]
    Last edited: Aug 31, 2006
  4. Sep 1, 2006 #3
    1)What are these formats?

    2)When evaluating the real integral, a contour integral is evaluated using the residue theorem over a contour only a part of which lies in the real line. The desired real integral is then found if the part of the contour lying outside the real line can be shown to vanish. How do I show this or at least recognise that this will be the case?

    3)Are you sure that integral is right? You seem to have missed some term, otherwise the integrand is a constant function.
    Last edited: Sep 1, 2006
  5. Sep 1, 2006 #4


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    Improper integrals of rational functions in which the denominator is different from zero for all real x and the degree of the denominator is at least 2 higher than the numerator. Like:

    [tex]\int_{-\infty}^{\infty} \frac{dx}{(1+x^2)^2}[/tex]


    The integral over the arc goes to zero since the absolute value of the integrand goes to zero faster than [itex]\pi R[/itex] goes to infinity. This is a consequence of the ML-inequality commonly used in CA to place upper bounds on contour integrals: Since the degree of the denominator is at least 2 higher than numerator, then:


    for sufficiently large R. Now, since you're integrating over a half-circle arc in these problems, the length of the contour is just [itex]\pi R[/itex] thus the integral over this region cannot be larger than:

    [tex]\left|\int_{C} f(z)dz\right|<\frac{k}{R^2} \pi R=\frac{k \pi}{R}[/tex]

    since [itex]|z|=R[/itex] on C.

    This goes to zero as R goes to infinity.

    Rational functions like the above, multiplied by Sin(x) and Cos(x) can also be evaluated by residue integration too like:


    Last edited: Sep 1, 2006
  6. Sep 2, 2006 #5
    Thanks saltydog for clearing it up. I will now try to evaluate the examples you gave. One more question, can residue calculus be usefull in integrations with finite limits and what would one do with the arc then?
  7. Sep 2, 2006 #6


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    Very good. Residue integration can be used for bounded integrands of sines and cosines. Like this:

    [tex]\int_0^{2\pi} Cos(\theta)d\theta[/tex]

    Make the substitution [itex]z=e^{i\theta}[/itex] noting that:


    Since [itex]\theta[/itex] makes one complete circuit, then z represents a circle around the (complex) origin thus the contour is closed and the integral becomes:


    The residue at the origin is zero so the integral is zero as expected. How about this one then:

    [tex]\int_0^{2\pi}Cos^2 \theta d\theta[/tex]

    That one's not zero. Can you prove it's [itex]\pi[/itex] using residue integration? Usually though this is done for more complex integrals like:


    Remember, only count the residues inside the contour.
    Last edited: Sep 2, 2006
  8. Sep 2, 2006 #7

    I'm not sure I understand this conversion into contour integral. Could you fill in some of the steps?

    Making the substitution [itex]z=e^{i\theta}[/itex], I get


    over the unit circle which seems to have a simple pole at the origin and a residue of 1, giving an integral of [itex]\pi i[/tex] using the residue theorem. What am I doing wrong?
    Last edited: Sep 2, 2006
  9. Sep 2, 2006 #8
    First I made the substitution [itex]z=e^{i\theta}[/itex], giving me

    [tex]\oint [\frac {z^4 + 1}{4z^2} + \frac {1}{2}]dz[/tex] around the unit circle. The contour integral of the constant term vanishes. Next I evaluate the residue at the second-order pole at origin using the formula

    [tex]Res(f, c) = lim_{z\rightarrow c} \frac {d^{n-1}}{dz^{n-1}} [f(z)(z-c)^n][/tex], getting 0 (which is surprising, can a residue at a pole be 0?).

    Using residue theorem, the integral evaluates to

    [tex]2\pi i * 0 = 0[/tex]

    Hard luck again. What went wrong?
  10. Sep 2, 2006 #9


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    If [itex]z=e^{i\theta}[/itex] then [itex]dz=ie^{i\theta}d\theta[/itex]

    You can generally deal with any rational function of sin's and cos's, this substitution will then yield a rational function of z. It's just a matter of locating the poles inside the unit circle.

    You should also keep in mind transformations that can get you into the correct form, like


    isn't an integral over a full circle with that substitution. How to fix this?

    yes it can. what's the residue of [itex]z^{-2}[/itex] at z=0?

    Do you have a complex analysis text? Ahlfors has a small section on the main common types. A book aimed more at the engineering crowd will probably be more example and problem rich though.
  11. Sep 3, 2006 #10


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    Thanks Shmoe. I think the differential of z caused the problem for Loom. It's just:


    so then:


    all the rest should fall into place. Also, I got most of the examples I gave above from the quintessential engineering math reference book: "Advanced Engineering Mathematics" by Kreyszig. It would certainly not do for math purist but . . . I ain't proud.
  12. Sep 3, 2006 #11
    Oh, I forgot to transform the differential form, how stupid of me.:">

    I don't have a complex analysis text, but I will be getting Spivak, which I think has a few chapters on complex analysis. Don't know whether it covers residues though.
  13. Sep 4, 2006 #12
    Got that one.

    Problem again, I get 2pi.

    After substitution, I have

    [tex]d\theta = \frac {1}{iz} dz[/tex]
    [tex]cos^2\theta = \frac{1}{4}(z^2+\frac{1}{z^2})+\frac {1}{2}[/tex]

    Making the integral

    [tex]I = \oint \frac {1}{iz}[\frac {1}{4}(z^2+\frac {1}{z^2})+\frac{1}{2}]dz
    =-i\oint \frac {z^4+2z^2+1}{4z^3} dz[/tex]

    [tex]f(z)=\frac {z^4+2z^2+1}{4z^3}[/tex]

    [tex]Res(f, 0) = lim_{z\rightarrow 0} \frac {d^2}{dz^2} (\frac{z^4+2z^2+1}{4z^3}z^3) = 1[/tex]

    [tex]I= -i * 2\pi i * 1 = 2\pi[/tex]

    What's wrong this time? Thanks for your patience.
    Last edited: Sep 4, 2006
  14. Sep 4, 2006 #13
    Substitute [itex]z=e^{i2\theta}[/itex]? However this gives a rather poor form of the integrand [itex]cos\theta = \frac{1}{2}(\sqrt{z}+\sqrt{\frac{1}{z}})[/itex], which doesn't seem easy to integrate using residues.

    Another possible method is to add a real portion from +1 to -1 to close the contour giving a half circle, in which case the function would have to be separately integrated over the real portion and the result substracted from the total integral to get the actual result. But this may be easier (being a rational function) than the original trigonometric one.

    But this one also wouldn't work because the pole would lie on the contour. Since the function will be defined almost everywhere on the real line, can I ignore the pole at the origin when integrating the real part and take an improper integral on both sides of the origin?
  15. Sep 4, 2006 #14


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    there's a problem with your residue calculation. you have:


    so the residue will be 1/2, it's the coefficient of the 1/z term in the Laurent series. Nothing fancy is needed to find the residue here, it's already a Laurent series.

    The "multiply by z^3, take second derivative" that you tried to do as you had a pole of order 3 should be "multiply by z^3/2!, take second derivative". There's a division by 1/(n-1)! to take into account the (n-1)! you get when you differentiate n-1 times. Remember how they derived this from the Laurent sereis?
  16. Sep 4, 2006 #15


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    You couldn't ignore this pole. If you wanted the upper half circle to be equal to an integral from [-1,1] (+ some residues) you need to have a closed contour. It would be possible to avoid this pole with a small half circle whose radius tends to 0 as you are taking your improper integral approaching this pole, but this is more work than is needed.



    and now you will get a full circle to integrate over.
  17. Sep 4, 2006 #16


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    You forgot the factoral coefficient in the calculation of a residue of higher order, you know, the :


    part. That would have one-halfed it down to 1/2 or . . . like Shmoe said, it's already in a Laurent series form with coefficient 1/2.
  18. Sep 5, 2006 #17
    Of course, it was already a Laurent series, and I forgot the factorial. No wonder I'm never good at math. Finally I've got this one. I'll move onto the next. Thanks to all.
  19. Sep 5, 2006 #18
    Are you using some result or theorem here? I haven't seen a result saying constants in the limits can be taken out of the integral. In fact I can see only a particular case where this would be possible, when the integrand had a linear primitive.
  20. Sep 5, 2006 #19


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    symmetry, [tex]\cos(\theta)=\cos(2\pi-\theta)[/tex]. The integral over 0,pi will be the same as the integral over pi, 2pi
  21. Sep 7, 2006 #20
    Hmm, I get the answer [tex]\frac{1}{2} \pi (4-3\sqrt {2})[/tex]. That's not the right answer, is it?
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