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Derivative of an integral

  1. Nov 1, 2012 #1
    1. The problem statement, all variables and given/known data


    Find d/dx[∫0x(cos(xt)/t)dt] (just in case the notation is not clear, the differential with respect to x, of the definite integral from x to 0, of cos(xt) over t)


    2. Relevant equations

    My initial thought was just using the fundamental theorem of calculus


    3. The attempt at a solution

    Let I(t)=∫0x(cos(xt)/t)dt

    Now, let f(t)=cos(xt)/t (the integrand) and let F(t) be an antiderivative of f(t) (a thought: f(t) is not defined at t=0, is this where I'm going wrong? Can I define F(t) to be an antiderivative? I mean, f(t) is not defined at t=0 but it's still integrable everywhere right?). Then by the fundamental theorem, I(t)=F(0)-F(x)

    So now we can take d/dx of both sides

    d/dx(I(t)) = d/dx[F(0) - F(x)]
    d/dx[∫0x(cos(xt)/t)dt] = f(0) - f(x) (by FTC)
    d/dx[∫0x(cos(xt)/t)dt] = cos(0)/0 - cos(x2)/x

    In this step I can't proceed because I'm dividing by zero. If I try to take the limit f(t) as t --> 0 the denominator blows up and the whole thing goes to infinity, so this limit does not exist. So it seems like this is an improper integral? I'm confused.


    ~~~~~~~~~~~~~```
    However, the book gives as the answer (1/x)(1-2cos(x2)) and I don't believe they use the FTC. I can use their method to just find the integral but I would like to see how to use the FTC in this case
     
    Last edited: Nov 1, 2012
  2. jcsd
  3. Nov 1, 2012 #2

    Mark44

    Staff: Mentor

    Here's the LaTeX version of the integral. You can click what I wrote to see how I did it.

    $$ \int_{t = x}^0 \frac{cos(xt) dt}{t}$$
    I would work with this integral, and take the limit of it as a approaches 0.
    $$ \int_{t = x}^a \frac{cos(xt) dt}{t}$$

    I haven't worked this through, but this is what I would try first.
     
  4. Nov 1, 2012 #3

    Zondrina

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    Just as mark said up there ^ I believe you'll be able to use the squeeze theorem afterwards to clean up you problem.
     
  5. Nov 1, 2012 #4
    Thanks guys. I am going to try this approach. Just out of curiosity, is there a reason I shouldn't apply the FTC in this problem? I mean, it appears its easier to do it this way that has been suggested to me in the responses, but is it still feasible to apply the FTC or is there a reason why that's not allowed in this case? I guess I'm asking, can I use the FTC when the integrand has a singularity at one of the end points of the interval I"m integrating over, or within the interval itself?
     
  6. Nov 1, 2012 #5

    Zondrina

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    The FTC allows you to evaluate the integral at the endpoints. The endpoints themselves are the cause of the problem, not the process in which you got to evaluating the endpoints. So yes you can still use the FTC, just pull the endpoint out as a limit so you can evaluate the integral properly and then check the limit after you're done.
     
  7. Nov 1, 2012 #6
    sorry for using attachment.I am working on my latex,hoping to be able to use it soon.
     

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  8. Nov 1, 2012 #7

    Dick

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    The complication that keeps you from simply using FTC is that you have x dependence in the integrand. You need to use http://en.wikipedia.org/wiki/Leibniz_integral_rule
    That will give you the book answer. That said, I think the exercise is a little silly since the integral does not exist for any nonzero value of x.
     
  9. Nov 1, 2012 #8
    for x and a greater than a positive number,there is no singularity in the interval of integration
     
  10. Nov 1, 2012 #9

    Dick

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    That's if you do a cutoff at a then let a->0. That lets you get a derivative that exists but the original integral (the thing that you are taking the derivative of) still doesn't exist. You do this sort of thing in physics, but it seems a little odd here.
     
  11. Nov 1, 2012 #10

    arildno

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    FTC is precisely what you use here.
    It's just that it uglifies, into Leibniz differentiation, due to its "essential" two-variable nature.
     
  12. Nov 1, 2012 #11
    this integral does converge,by Dirichle criterion.Examine uniform convergence for using Leibnitz formula.
     
  13. Nov 1, 2012 #12

    Dick

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    I'm not sure I believe that. Put x=1. The resulting integral doesn't look convergent me by a simple comparison test.
     
  14. Nov 1, 2012 #13

    HallsofIvy

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    The general Leibnitz formula says that
    [tex]\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x, t)dt= \frac{d\beta(x)}{dx}f(x,\beta(x))- \frac{d\alpha(x)}{dx}f(x,\alpha(x))+ \int_{\alpha(x)}^{\beta(x)}\frac{\partial f(x,t)}{\partial x}dx[/tex]
     
  15. Nov 1, 2012 #14

    arildno

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    It is, most definitely, convergent
     
  16. Nov 1, 2012 #15
    Thank you guys I am finding this discussing very useful and hedipaldi, thank you so much for taking the time to upload that document. This is all so appreciated by me!
     
  17. Nov 1, 2012 #16

    Dick

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    The formula for the derivative is convergent, if that's what you mean. The integral you are taking the derivative of isn't. Most definitely.
     
    Last edited: Nov 1, 2012
  18. Nov 2, 2012 #17
    yes,i agree.
     
  19. Nov 2, 2012 #18
    the initial integral converges al the infinity but not at 0,so the differentiation formula is valid only when the interval does not contain 0
     
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