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Evaluate (1-x^3)^-1/3 from 0 to 1

  1. May 31, 2009 #1
    I think I understand how to get the answer from the AoPS thread, but i dont understand how we can divide it up into three sheets. Why is this possible? Can someone explain why also this doesnt occur with branch points that are real, i.e. the cut is only on the real axis? also, how does one decide which branch of the residue at infinity to use? i know it must be exp(-5pi/3) because thats the only real answer when finally doing out the arithmetic at the end of the problem, but what about the general case? could it be possible to have more thn one real answer, Again, if so, how do you choose? Thank you so much for any help at all. If you dont know what I am talking about, go to forums at art of problem solving and search mercedes benz contour. That should pop up the thread. Thanks again!
  2. jcsd
  3. Jun 2, 2009 #2
    I think it would be instructive for you to do the integral in a slightly different way. In the way you have seen it, it suggests that the choice of the branch cuts is critical. Another way to compute the integral is to combine the Mercedes Benz contour with a circle of radius R. Join the upper half of both contours into one contour and the lower halfs of both into another contour.

    You take the branch cut of the function you integrate over the contour in the upper half plane to be in the lower half plane. It doesn't matter exactly in what direction they run, as long as they don't intersect the contour in the upper half plane. Then you define the function in the lower half plane by analytical continuation of the function in the upper half plane. The branch cut of this function must be chosen to by in the upper half plane, but the exact details are irrelevant.

    What you then see happening is that when you cross the branch cut of the other function, the function value picks ups a phase factor relative to the other function. Then, because going one round around the origin doesn't lead to a net phase factor, you can combine the integrals over the two halfs of the big circle into a single integral of an analytical function. By applying the transformation z ---> 1/z you can apply the residue theorem.
  4. Jun 2, 2009 #3
    yes but wouldnt theta have to be restricted -2pi/3, 4pi/3 in order for it to be discontinuous along the cut? I always get the answer to be having a 9 in the denominator. How does one go about calculating the lmits above and below the cut when the cut is at an angle, i.e. not real?
  5. Jun 3, 2009 #4
    Consider the contours:

    1) Integral over the real axis from zero to 1-epsilon, half circle to 1 + epsilon, then onwards to z = R and then a third of a circle counterclockwise to exp(2/3 pi i) R, then from there to
    exp(2/3 pi i) [1 + epsilon], then half a clockwise circle to
    exp(2/3 pi i) [1 - epsilon], and then back to the zero.

    2) Take contour 1) and rotate it by 120 degrees, i.e. multiply all the points by exp(2 pi i/3)

    3) Rotate contour 2) again by 120 degrees.

    Contours 1), 2) and 3) put together form the single MB contour you've been looking at. If you want to do the problem on this single contour, you have to get the branch cuts exactly right so that the discontinuity is on the right places.

    But it it much easier to consider the three parts separately and for each contour just take the branch cuts so that they don't intersect the contour in any arbitrary way you like. The functions on the different contours are taken to be analytical continuations of each other.
  6. Jun 3, 2009 #5
    If you define the polar cordinates such that the integral along the real axis from zero to 1 is equal to the real integral we want to calculate and call that I, then we find that the three contours are given by:

    I[1-exp(2 pi i/3)] + IR +IB

    where IR is the circle segment of the circle with radius R and IB are the segments from radius 1 to R that will cancel between the 3 different contour integrals. So, the sum of the 3 contour integrals is:

    3 I[1-exp(2 pi i/3)] + I_circ

    Where I_circ is the integral over the big circle of radius R. Since all three individual contour integrals are zero by the residue theorem, we have:

    3 I[1-exp(2 pi i/3)] + I_circ = 0

    I_circ is the integral over the circle of (1-z^3)^(-1/3). From the way we defined the polar coordinates, we see that this is the same as the integral of exp(pi i/3) (z^3 - 1)^(-1/3), because to get the real integral I when integrating from zero to 1 minus epsilon, implies that for z between zero to 1 minus epsilon, the polar angle of 1-z was taken to be zero, so the polar angle theta of 1-z has to be chosen to be minus pi between z = 1 +epsilon and infinity. The integral is thus given by:

    -2/3 pi i exp(pi i/3) /[1-exp(2 pi i/3)] =

    2 pi sqrt(3)/9
  7. Jun 3, 2009 #6
    i guess what I want to ask is this:

    How can we find the phase of each segment? How would one calculate it to find that each one does indeed equal I or exp(2pi*i/3)I? Can you tell me what interval theta is restricted for each branch point in local polar coordinates, so that I might find the limits on each side of the segment for myself?
  8. Jun 3, 2009 #7
    On each of the contours we have to define the function

    f(z) = (1-z^3)^(-1/3)

    We have:

    1-z^3 = [1-z][exp(2 pi i/3) - z][exp(4 pi i/3) - z]

    For the first contour, we need to choose the branch cuts so that they don't intersect the first contour (but they may intersect the other contours). So, we can choose the branch cut at z = 1 to point downward, at z = exp(2 pi i/3) to point to the left and at z = exp(4 pi i/3) to point downward. You don't have to bother about the precise details.

    Then we need to define the polar angles to fix f(z) unambiguosly on the first contour. Let's denote the definition of f(z) that we'll choose by f1(z). Consider the polar coordinates for 1-z, exp(2 pi i/3) - z, and
    exp(4 pi i/3) - z. Let's make this (arbitrary) choice:

    For 1 - z we take the polar angle to be zero for real z smaller than 1.

    For exp(2 pi i/3) - z we take the polar angle to be 2 pi/3 for
    z = r exp(2 pi i/3) and r<1

    For exp(4 pi i/3) - z we take the polar angle to be 4 pi/3 for
    z = r exp(4 pi i/3) and r<1

    This choice makes f1(z) real on the real axis between zero and 1, but note that you could have made any arbitrary choice. Now, with the polar angles defined in particular directions and the branch cuts chosen, f1(z) is completely fixed.

    The integral from zero to 1 is then the desired integral I.

    The integral from exp(2 pi i/3) to zero is minus the integral from zero to exp(2 pi i/3). If you substitute z = x exp(i pi i/3), then you see that the integral becomes minus exp(2 pi i/3) I. Note that it follows from the way we defined f1(z) that f1[r exp(2 pi i/3)] = f1(r) for real r < 1.

    On contour 2, we define f2(z) by taking f1(z) and analytically continuing it from points r exp(2 pi i/3) for r > 1. This ensures that on the big circle with radius R, we end up integrating a single analytical function. This means that the polar angles are completely fixed, as they are continued in a continuous way from the way f1(z) was defined. The branch cuts for f2 are chosen such that they don't intersect contour 2.

    It is then easy to see that the integral from zero to exp(2 pi i/3) is I. To see this, consider f2[r exp(2 pi i/3)] for r < 1. Since f2 is analytically continued from f1 for r > 1, the polar angle of
    exp(2 pi i/3)-z for r < 1 is 2 pi. This is because the polar angle has to agree with the way f1 was defined for r > 1 and if you walk around the branch point anti-clockwise to arrive at r < 1, you cannot encounter any branch cuts, so the polar angle will increase to 2 pi. The branch cut must be chosen so as not to intersect contour 2, so it could be chosen to point into contour 1.

    This polar angle of 2 pi leads to the integral of f2 picking up a phase factor of [exp(2 pi i)]^(-1/3) = exp(-2 pi i/3) relative to what the integral of f1 over that segment would have been. In fact, if you think about this, the rule for the phase factor should be obvious: every time the new analytically continued function f2 crosses the branch cut of the old function f1 you get a phase factor of exp(-2 pi i/3) because the polar angle must have jumped by 2 pi relative to the old function (the new function is continuous on contour 2, the old one will have a branch cut somewhere).

    Then exp(-2 pi i/3) times exp(2 pi i/3) I = I.

    You then easily see that over contour 2, you get essentially the same thing as over contour 1.
  9. Jun 5, 2009 #8
    thank you! much appreciated
  10. Jun 7, 2009 #9
    On the last part of the qoute, wouldnt the angle only increase by 2pi if theta is restricted to -pi/3, 5pi/3, making the cut not crossable on the clockwise direction, so you have to go all the way around it from behind to gedt 2pi? If not, why?
  11. Jun 7, 2009 #10
  12. Jun 8, 2009 #11
    why is it that it must gain a 2pi in the phase? is it because we defined a "cut" on the real axis in between 1 and negative infinity and in the norm , i.e. exp(2p*i/3) - xexp(2pi/3) becomes 1-x? This way the argument is the same for the 1-z factor. Also, when you mean we could have made any arbirtrary choice, do you mean so long as the two polar angles cancel for the 2pi/3 factor and the -2pi/3 factor cancel? I just need a little clarification on this method, you are helping me very much.
  13. Jun 8, 2009 #12
    also, if we have the branch cuts point in any which way, doesnt that contradict the residue theorem, i.e. it must be analytic inside the contour except for a finite number of singularities?
  14. Jun 8, 2009 #13
    Ok, let's summarize things a bit (I'll give more detailed replies later). There are two methods:

    1) You integrate over a single contour.

    2) You chop the contour into 3 separate contours and on each of the contoiurs you are integrating a different analytical function

    In 1) you don't have much freedom to choose the branch cut. You do have a freedom left to choose the polar angles.

    In 2) you can choose the branch cuts of the three functions differently.

    To answer your last question about the residue theorem: Yes, because the function has branch point singularities inside the contour, you cannot apply the residue theorem in the usual way.

    But what you can do is perform a conformal transformation
    z ----> 1/z. Under this transformation, the region outside the contour moves inside the contour and what was inside goes outside. The contour is then traversed in the opposite direction but you also have:

    dz ---> -1/z^2 dz

    So, we can omit the minus sign and traverse the tranformed contour in the same direction and integrate the function f(1/z)1/z^2 dz. The branch cut singularities have moved outside and inside there are no singularities except for a pole at z = 0, which corresponds to a pole at infinity before the conformal transformation was applied. So, we can now apply the residue theorem.

    Instead of using the transformation z ---> 1/z, you can also connect the contour to a big circle (you have to do that if you use method 2) such that the total contour's interior is what is inbetween the circle and the inner part of the contour, so the branch points are outside that contour. You can then simply evaluate the integral over the big corcle in the limit that the radius goes to infinity. That then amounts to the same thing as applying the transformation z --> 1/z and consider infinitessimal radius, which in turn is the same as considering the residue at the pole at z = 0.
  15. Jun 10, 2009 #14
    ok, Ill await you future posts
  16. Jun 11, 2009 #15
    Let's start again with method 1. And let's not care about the phase factors at all, as that then makes clear that you really have the freedom to make arbitrary choices here.

    There one important point:

    The function we integrate over the contour must not have any discontinuities on the contour. It doesn't have to be analytic inside the contour, and even the singularities inside the contour don't have to be poles, they can be branch cut singularities.

    We can then make sure the function we integrate over is analytic in some neighborhood of the contour and then try to apply the residue theorem in some way.

    Then, this means that we have to put the branch cuts along the lines from the origin to the three branch points. Also, we need to bend the branch cuts coming from from exp(2 pi i/3) and exp(-2 pi i/3) to the negative real axis after they arrive at the origin, because that way you have combined the three branch cut into one that yields a net phase factor of exp(2 pi i) when crossed. The branch cuts then don't cause a discontinuity on the contour.

    Within this constraint we have to choose a defintion for the function

    f(z) = (1-z^3)^(-1/3)

    everywhere on the complex plane. We already made a choice for the branch cuts. So, we have to define the polar coordinates relative to the branch points of the factors z_b - z. If we fix these for one particular value of z, then it is completely fixed everywhere. Let's take z to be very close to the origin, just below the positive real axis. Then z_b - z is simply z_b itself if we deform the branch cuts slightly so that the branch cut from z = 1 passes above the origin while the branch cuts that come from the two other branch cuts bend to the negative real axis just before they reach the origin.

    So, all we need to do is simply assign polar coordinates to the branch points in the form of
    r exp(i theta). The most general choice is:

    z_b1 = exp(2 pi i r)

    z_b2 = exp(2 pi i/3) exp(2 pi i s)

    z_b3 = exp(-2 pi i/3) exp(2 pi i t)

    Where r, s and t are arbitrary integers that we don't need to fix.

    We then have that

    f(z) = (zb1-z)^(-1/3) (zb2-z)^(-1/3) (zb2-z)^(-1/3)

    here it is to be understood that each z_bi-z is assigned polar coordinates, such that for z = 0 (and with deformed brach cuts) we have the above expressions in terms of polar coordinates. Then the choice of the branch cuts we have made, fixes what the polar representation of z_bi - z should be everywhere else on the complex plane. And, by definition we have

    [r exp(i theta)]^p = r^p exp(i theta p)

    so f(z) is now unambiguously defined.

    Let's now compute the contour integral. From zero to 1, infinitessimally below the branch cut on the real axis if we had fixed r = s = t = 0, we would have had:

    f(x) = (1-x^3)^(-1/3)

    The (z_b1 - z)^(-1/3) factor has an extra phase factor of exp(-2 pi i/3 r),

    The (z_b2 - z)^(-1/3) factor has an extra phase factor of exp(-2 pi i/3 s),

    The (z_b3 - z)^(-1/3) factor in here has an extra phase factor of exp(-2 pi i/3 t),

    so we have:

    f(x)= (1-x^3)^(-1/3) exp[-2 pi i/3 (r+s+t)]

    When we are infinitessimally above the real axis between zero and 1 the (z_b1 - z)^(-1/3) has an extra phase factor of exp(-2 pi i/3). So we see that he integral from zero to 1 infinitessimally below the real axis and then from one to zero infinitessimally above te real axis is:

    [1-exp(-2 pi i/3)] exp[-2 pi i/3 (r+s+t)] I

    where I is the real integral you want to compute.

    Now, you can easily find the integrals along the two other branch cuts from zero to z_b2 and z_b3 as follows. Note that that just above the real axis we have:

    f+(x) = (1-x^3)^(-1/3) exp[-2 pi i/3 (r+s+t)] exp(-2 pi i/3)

    Then just to the right of the branch cut from zero to z_b2 at the point z = z_b2 x, each of the factors (z_bi - z)^(1/3) gains a phase factor of exp(-2 pi i/3), so the function there is:

    f(z_b2 x) = f+(x)

    On the left side of the branch cut the function is f(x) exp(-2 pi i/3) So, the integral from zero to z_b2 and back via the other side is:

    Integral from zero to z_b2 on the right side f(z)dz - Integral from zero to z_b2 on the left side f(z)dz =

    put z = z_b2 x and integrate from x = 0 to 1

    Integral from zero to 1 of f(z_b2 x) z_b2 dx =

    Integral from zero to 1 of f+(x) z_b2 dx = exp[-2 pi i/3 (r+s+t)] I

    On the ay back we get the same multiplied by minus exp(-2 pi i/3), so we see that the integral is the same as the itegral along the two sides of the segment from zero to 1 on the real axis:

    [1-exp(-2 pi i/3)] exp[-2 pi i/3 (r+s+t)] I

    Similarly you can easily see that the third part of the contour also gives the same result.

    Then you need to compute the residue at infinity. If you do that, you see that the factor
    exp[-2 pi i/3 (r+s+t)] will pop up there and you will get a result for I that doesn't depend on r, s and t.
  17. Jun 12, 2009 #16
    when you say when the function is infinitesimally above the real axis, I know why its true that it gaisn a phase factor of -2pi/3 but i dont fully understand this. can you help me
  18. Jun 12, 2009 #17

    When you are just to the left of the branch point at z = 1 and you go from below the real axis to above the real axis by swinging around the branch cut at z = 1, the polar angle increases from exp(2 pi i r)
    to exp(2 pi i r)*exp(2 pi i).

    This means that the factor (z_b1 - z)^(-1/3) gains a phase factor of

    [exp(2 pi i)]^(-1/3) = exp(-2 pi i/3)

    I now see that I made a mistake in explaining why along the other branch cut in the direction of z_b2 the function f(z_b2 x) is the same as f(x), I'll correct that later today.
  19. Jun 12, 2009 #18
    I think I understand now, but why is it that the 2pi is on the bottom of the cut if we take the function sqrtx and have the cut to be the positive real axis as opposed to the top?
  20. Jun 12, 2009 #19
    also, why isnt 8pi/3 the limit on the left side of the cut going to the left at an angle on the bottom? It makes more sense for it to be on the bottom because when we rotate theta anticlockwise theta the cut is hit by theta at 2pi/3, the going 2pi again would bring 8pi/3 on the left , but when i chekc to see if the cut on the negative real axis cancels, this si not the case so this cannot be, but it does cancel when the limits are the reverse of what im talking about.
  21. Jun 13, 2009 #20
    First, let me correct the mistake I made when arguing that

    f(z_b2 x) = f+(x)

    What happens is that the factors (z_b3 - z)^(-1/3) and
    (z-z_b1)^(-1/3) taken together on the line from the origin to z_b2 just to the right of the branch cut have the same phase factor as at the origin. just above the branch cut on the real axis. Due to to symmetry if one factor has an extra phase factor of exp(i theta), the other has an extra phase factor of exp(- i theta).
  22. Jun 13, 2009 #21
    This is just an arbitrary choice. If you do it differently, the contour integral will pick up a minus sign relastive to the old convention, and the relation between the real integral involving the (positive) square root and the contour integral will now also involve an extra minus sign.
  23. Jun 13, 2009 #22
    how does the real integral get an extra minus sign though? I'm seeing that these minuses are supposed to cancel out
  24. Jun 13, 2009 #23
    Well, what happens is that if you choose the other convention, the integral slightly above the real axis will be minus the real integral if it previously was the real integral. The value of the contour integral will be minus what it was (the residue theorem will yield an extra minus sign). So, the result for the real integral will be the same.
  25. Jun 13, 2009 #24
    yes but why is it the the residue get the extra minus sign?
  26. Jun 13, 2009 #25
    residue theorem*
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