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Integration by substitution: Can I treat this as constant

  1. May 28, 2014 #1
    I am trying to compute the following integral:

    [itex]\int \exp^{w^T \Lambda w}\, d\theta[/itex] where [itex]\Lambda[/itex] is a constant wrt [itex]\theta[/itex]

    [itex]w = y - t(x, \theta)[/itex]

    So, I am trying to use substitution and I have:

    [itex]d\theta = \frac{-dw}{t^{'}(x, \theta)}[/itex]

    So, substituting it, I have the following integral to compute:

    [itex]\int \frac{exp^{w^T \Lambda w}\, dw}{t^{'}(x, \theta)}[/itex]

    Can I treat [itex]t^{'}(x, \theta)[/itex] as a constant? My instinct tells me no as there is a relationship between [itex]w[/itex] and [itex]\theta[/itex] given by this function [itex]t[/itex], but I just wanted to make sure.
     
  2. jcsd
  3. May 28, 2014 #2
    I don't think so. As you can see t'(x,θ) still depends on θ, while your integration variable is now w. What you need to do is to invert the relation and express everything as a function of w. In general this won't be a constant.
     
  4. May 28, 2014 #3
    Thank you. That is what i thought. Unfortunately, it is not easy to invert this function. I will need to think in terms of linearisation with Taylor series or something...

    Thanks!
     
  5. May 28, 2014 #4
    Be careful when using Taylor series inside an integral. Keep in mind that you are integrating over all the possible values of θ and so you can never truncate a series in θ. What you can do is, for example, Taylor expand in x, if it is small, since you are not integrating over it.
     
  6. May 28, 2014 #5
    Thanks for that. However, at least in this problem, x is a constant. So, I am interested in expansion in the [itex]\theta[/itex]. However, I thought this was the standard way to deal with such non-linear terms, otherwise I am not sure what can be done. As long as I am expanding close to the point of interest, I thought I should be able to avoid these higher order terms without the accuracy suffer too much?
     
  7. May 28, 2014 #6
    It strongly depends on what are your boundaries for the integration. If for example, θ is integrated from -∞ to +∞ then you can't really say what higher order terms are negligible since θ range from arbitrarily small values to arbitrarily large ones.
    On the other hand if, for example, θ is only integrated over a small region around zero, depending on how large is this region, you could maybe expand around small values of θ.

    I recommend extra caution when doing Taylor expansion inside an integral.
     
  8. May 28, 2014 #7
    That is a good point. Thank you for pointing that out. I will need to think about this.
     
  9. May 28, 2014 #8
    I was just reading a bit about this quadrature based integration and I was wondering if that can somehow be applied to this case? I am afraid I do not know much about it but was wondering if this is something worth looking into.

    Any suggestions on how to approach this problem would be very appreciated!
     
  10. May 28, 2014 #9
    From your post I don't understand if you have an explicit expression for t. If so, what is that?
     
  11. May 28, 2014 #10
    [itex]t[/itex] is some non-linear mapping between the space of x and y. They do not necessarily have a parametric form though I can probably find some parametric representation for this mapping in terms of splines or something. However, in general I do not have an explicit expression for it.

    Just to give a background, I am working in image analysis and x and y are observed images. t is the non-linear mapping that takes one from space of x to the space of y.
     
  12. May 28, 2014 #11
    Well, if you don't know what the expression for t is then it's pretty hard to tell whether or not you can use a certain method to approximate your integral or not. The most suitable method could depend, for example, if t' is varying fast or slowly, if it is peaked at some particular value and so on. Without any such information you can't clearly solve the integral.
     
  13. May 28, 2014 #12
    Yes, typically each 3D point in the space of image x is moving to another 3D point in the space of image y. The problem is regularised in some way but the function is still highly non linear with many local minima possible and quite severely under-determined. So, I guess I am a bit stuck here.
     
  14. Jun 2, 2014 #13
    I was wondering if I can assume that [itex]t[/itex] varies slowly, if this could help with this integration in some way. So, I can assume that [itex]t[/itex] is a smoothly changing function and can be differentiated infinitely even. Would that help in solving this integral?
     
  15. Jun 2, 2014 #14
    What I honestly don't understand is what you mean by "solving the integral". Since you don't exactly know the expression for t you can't solve the integral, you will alway have to figure out what t is, even if you can use some of its properties.
     
  16. Jun 2, 2014 #15
    I will start a new thread about this as things will get confused perhaps. I will try and generate an expression for t using some interpolation function.
     
  17. Jun 2, 2014 #16
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