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Substitution of imaginary variables in integral?

  1. Sep 14, 2013 #1
    I wanted to do this integral $$\int_a^b \frac{dx}{1-x^2} $$ and I was able to get the right answer with the substitution u=ix, where i is the square root of -1.

    But is this a valid mathematical procedure? $$\int_a^b \frac{dx}{1-x^2}=i \int_{-ia}^{-ib} \frac{du}{1+u^2}$$

    Do those limits even make sense? They don't make sense in terms of area under a curve. But the integral over real numbers is inverse tangent, and if you just plug in the imaginary number into inverse tangent, you get the right answer?
     
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  3. Sep 14, 2013 #2

    fzero

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    I think the substitution is fine as long as the variable is pure imaginary, because the analysis is equivalent to that on the real line. Computing an arctan of an imaginary number has to be done carefully, since the function has branches. You could also do the integral without substitution by partial fraction decomposition in terms of ##1/(i \pm x)##, in which case you will also consider the branches of the log function.
     
  4. Sep 15, 2013 #3

    arildno

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    Yes, the procedure is valid, with caveats already mentioned by fzero.

    Switching integration to some path in the complex plane is a nifty trick in "advanced" integration theory (you can get radical simplifications in the computational burden), but you have to know what you are doing...
     
  5. Sep 15, 2013 #4
    I really think Alice says it best in regards to applying the Fundamental Theorem of Calculus to line integrals over multi-valued functions and feel she should be quoted in all new Complex Analysis textbooks:

    https://www.physicsforums.com/showthread.php?t=702808
     
  6. Sep 15, 2013 #5

    arildno

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    :smile:
     
  7. Sep 15, 2013 #6
    Thanks everyone. I'm slightly confused though. I'm a little bit familiar with complex integration. However, is this a case of complex integration? The original integral is on the real number line. Is the new integral with limits -ia and -ib on the imaginary line? It can't be, because in order to get it there, don't you have to deform the contour so that it goes over the imaginary line? But not such deformation took place, just a substitution of variables?
     
  8. Sep 15, 2013 #7
    Ok, first off I get a different integral if I let [itex]u=ix[/itex]:

    [tex]\int_a^b \frac{1}{1-x^2}dx=-i\int_{ia}^{ib} \frac{1}{1+u^2}du, \quad u=ix[/tex]

    Secondly, they're all contour integrals, the real ones are just real integrands over the real axis. So take a "contour" over the real axis from a to b and then make the transformation [itex]u=ix[/itex] for x over that path and we create a new path along the imaginary axis from ia to ib.

    And finally, most importantly, if we map that path along the imaginary axis onto the antiderivative in an analytically-continuous fashion, then we can write:

    [tex]\int_a^b \frac{1}{1-x^2}dx=-i\int_{ia}^{ib} \frac{1}{1+u^2}du=-i\arctan(u)\biggr|_{ia}^{ib}[/tex]

    I should point out if the integration path includes the point i or -i, then we can't simply map the path onto the arctan function using a straight line. Passing through those points would not be in an analytically-continuous fashion. We'd have to indent around those singular points.
     
    Last edited: Sep 15, 2013
  9. Sep 15, 2013 #8
    I think I got confused because often you see problems where an integral over I1 can be written as an integral over I2 if I1 +I2 form a closed curve and no poles are enclosed. In that case the contours I1 and I2 are in the same complex plane.

    But I guess you're saying that a substitution involves a mapping from one complex plane to another. So [a,b] on the real line in the complex plane z gets mapped to [ia,ib] on the imaginary line of complex plane u, where z and u are different planes.

    And here I guess you mean we would have to deform the contour in the same complex plane, and not use a substitution of variables?

    Is it true that you can always perform the substitution u=ix no matter if your function is well-behaved or not, but only when using the fundamental theorem of calculus do you have to worry about analytic-continuity? So for

    [tex]\int_a^b \frac{1}{1-x^2}dx=-i\int_{ia}^{ib} \frac{1}{1+u^2}du=-i\arctan(u)\biggr|_{ia}^{ib}[/tex]

    the first equality is always true, but the second may or may not be true depending on analycity?
     
    Last edited: Sep 15, 2013
  10. Sep 15, 2013 #9
    That's deformation of path when the integrand is analytic in a region containing both paths. That's not relevant here.

    That's exactly right: we have a path in the (complex) x-plane and a mapping, u=ix, taking the path into another path in the complex u-plane.

    I think you're referring to my comment about "indenting" around the singular points of arctan(u) which are at u=i and u=-i. We could do that for example if we wanted to compute the principal-valued integral:

    [tex]\text{P.V.} \int_0^2 \frac{1}{1-x^2}dx[/tex]


    As long as the path is analytically-continuous (smooth and differentiable) over the antiderivative, then we can use it's values at the endpoints to compute the integral. In the case of the principal-valued integral above, if we were to make the substitution u=ix and integrate over u from 0 to 2i and indent around i, we loose analyticity by creating a half-circle contour connected to two straight lines and so must break-up the integral into three integrals, two over the straight paths, and one over the half-circle if we wish to use antiderivatives to compute the integral.
     
  11. Sep 16, 2013 #10
    I wish to state for the record that I believe this is not correct: it has nothing to do with indenting the path but rather I believe since the original path is going through a singular point, we cannot use a single antiderivative for this computation. For one thing, we could just deform the indentation until it were smooth at the junction of the straight line segments. And I can certainly construct a bent contour over an analytic domain of an antiderivative and compute the associated integral with the end-point values from a single antiderivative, that is, as long as the contour does not pass through a singular point rendering the path non-analytic.
     
    Last edited: Sep 16, 2013
  12. Sep 16, 2013 #11

    vanhees71

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    This is not such an easy issue. I'd stay within real analysis for this integral to be on the save side.

    First of all it's clear that one shouldn't have 1 and -1 in the (closed!) intervall [itex][a,b][/itex].

    Now let's first look for the antiderivative of the integrand, which is most easily found using
    [tex]\frac{1}{1-x^2}=\frac{1}{2} \left (\frac{1}{1+x}-\frac{1}{1-x} \right ).[/tex]
    This gives
    [tex]F(x)=\int \mathrm{d} x \frac{1}{1-x^2} = \frac{1}{2}[\ln(|1+x|-\ln(|1-x|)], \quad x \neq \pm 1.[/tex]

    Second we have to distinguish three cases:

    (a) [itex][a,b] \subset (-1,1)[/itex]
    (b) [itex][a,b] \subset (1,\infty)[/itex]
    (c) [itex][a,b,] \subset (-\infty,-1)[/itex]

    For case (a) we have
    [tex]F(x)=\frac{1}{2} [\ln(1+x) - \ln(1-x)]=\frac{1}{2} \ln \left (\frac{1+x}{1-x} \right ) = \mathrm{artanh} \; x[/tex]
    and thus
    [tex]\int_a^b \mathrm{d} x \frac{1}{1-x^2}=\mathrm{artanh} \; b-\mathrm{artanh} \; a.[/tex]

    For case (b) we have
    [tex]F(x)=\frac{1}{2} [\ln(1+x)-\ln(x-1)]=\frac{1}{2} \ln \left (\frac{x+1}{x-1} \right )=\mathrm{arcoth} \; x[/tex]
    and thus
    [tex]\int_a^b \mathrm{d} x \frac{1}{1-x^2}=\mathrm{arcoth} \; b-\mathrm{arcoth} \; a.[/tex]

    For case (c) we have
    [tex]F(x)=\frac{1}{2} [\ln(-x-1)-\ln(1-x)]=\frac{1}{2} \ln \left (\frac{x+1}{x-1} \right ) = \mathrm{arcoth} \; x[/tex]
    and thus
    [tex]\int_a^b \mathrm{d} x \frac{1}{1-x^2}=\mathrm{arcoth} \; b-\mathrm{arcoth} \; a.[/tex]

    Using complex integration is pretty subtle here, because you have logs involved which have a branching point at its arguments vanishing. Thus integrating in the sense of complex functions, you have to be very careful about choosing your path of integration, connecting [itex]a[/itex] and [/itex]b[/itex] on the real axis. It depends on, how you deform your real integration path. This is the more true, if [itex]1 \in (a,b)[/itex] or [itex]-1 \in (a,b)[/itex]. Then the real integral as it stands is undefined. Sometimes in applications, you have to make sense of such an integral, and then the context of your application decides in which sense you have to read the integral. You can define it by deforming the originally real integration path in the complex plane, but the result depends on, which path is chosen, i.e., which branches of the logarithms you take and where you cut the plane for to define the particular Riemann sheet you choose your logarithms from.
     
    Last edited: Sep 16, 2013
  13. Sep 17, 2013 #12
    Oh vanhees you take all the fun out of it.

    There is a saying in martial arts, "cry in the dojo, laugh on the battle field". Well, we're all in the math dojo in here, learning from all the great masters of the world, struggling through kata, learning all the ancient techniques and depending how much we practice will in a large part determine our success in battle and if we stay in math long enough we will one day confront problems that no one else on earth can help us solve and will have to battle them alone.

    Now I argue that we can solve the integral:

    [tex]\text{P.V}\int_0^2 \frac{1}{1-x^2}=-i\int_{0}^{2i} \frac{1}{1+u^2}du[/tex]

    using antiderivatives in such a way that the answer does not depend on how we indent around the singular point and agrees with the real analytic analysis. I'm not sure though as I haven't worked it out. But when you're in the dojo, you can't be afraid to fight. :)
     
  14. Sep 17, 2013 #13

    vanhees71

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    Well, writing the principle-value symbol in front of the integral makes everything well defined, but I still don't like the idea to unnecessarily introduce complex integration here. Just use the definition.

    Writing again
    [tex]\frac{1}{1-x^2}=\frac{1}{2} \left (\frac{1}{1+x}-\frac{1}{1-x} \right),[/tex]
    we need to worry about the principal value only for the second term since only for it the singular point [itex]x=1[/itex] is within the integration region. Thus we only need
    [tex]\mathrm{PV}\int_0^2 \mathrm{d} x \frac{1}{x-1}=0,[/tex]
    because the integrand is odd at [itex]x=1[/itex] and the integration interval symmetric around this point.

    One should not fight with complicated weapons if there is a simpler way ;-).
     
  15. Sep 17, 2013 #14
    . . . oh. I didn't see that way. Nice! Thank you.


    Agreed.

    Still though I may wish to pursue the complex way just for fun to see if it does indeed not matter how we indent around the singular point.
     
  16. Sep 17, 2013 #15

    vanhees71

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    Sure, it's a great thing to play with the integral, using complex integration. You may use different paths to define the various branches of the integral, determine which paths lead to the same and which to different results, etc. This is of course a more advanced issue than the original question on the real integral!
     
  17. Sep 17, 2013 #16
    Gonna' have to disagree with you vanhees not that I want the mentors to get involved and lock the thread but he did say integrate from "a" to "b" without specifying what those values were and I think you and I would both agree if that were a test question and we constrained a and b to just baby values like from 1/2 to 3/4, then I'm sure our professor would pounce on us for not considering all cases. :)

    That is, the integral could be complex if a and b were, not that we need to use complex means. As well as the principal-valued case which I think would catch a lot of students.

    And one final comment regarding the OP's comment:

    An integral of a complex-valued function IS area under a curve. Oh, that would be a good test question guys to see if you know what's going on: Explain how the complex integral:

    [tex]\int_{i/4}^{3i/4} \frac{1}{1+z^2}dz[/tex]

    is (real) area under a curve. Just for fun if you want. :)
     
    Last edited: Sep 17, 2013
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