1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Showing an integral converges

  1. Nov 12, 2013 #1
    1. The problem statement, all variables and given/known data
    Let g: R→R be an infinitely differentiable function satisfying g(x)=0[itex]\forall[/itex] |x|>1

    Consider the integral I[itex]\lambda[/itex] = ∫g(x)sin([itex]\lambda[/itex]x2) dx taken from -∞ to +∞

    Prove that I[itex]\lambda[/itex]→0 and identify the rate of decay as [itex]\lambda[/itex]→0

    I have no idea how to start this. I thought maybe differentiation under the integral or using Bessel's inequality?
  2. jcsd
  3. Nov 12, 2013 #2
    I can get you started. First, since g(x) = 0 for |x|>1, you are integrating only from -1 to 1. We also see that as ##\lambda \rightarrow 0## sin##(\lambda x^2) \rightarrow 0##. If g(x) is infinitely differentiable, can you show it is bounded on [-1,1]? And then what happens to that integral as ##\lambda \rightarrow 0##?
  4. Nov 12, 2013 #3
    Alright, I think I see where you're going. I can show it's bounded by setting m=min g on [-1, 1] and M=max g on [-1, 1] so that m*I[itex]\lambda[/itex]<I[itex]\lambda[/itex]<M*I[itex]\lambda[/itex].

    It seems like I'd do a series expansion about [itex]\lambda[/itex]=∞, but I have no idea what the expansion would look due to the nature of g
  5. Nov 12, 2013 #4
    Your bounds are correct, but not quite to the point here. The boundedness of g tells us that as sin(##\lambda x^2##) sinks down to 0, the g won't pull it back up and ruin our limit.

    Also, you can't just say g is bounded, you have to give a reason. What is it?

    In terms of the decay, I wasn't sure what derivative we need to look at (not much of a physicist). I suppose it is d##\lambda##/d??? but I don't know what the ??? would be. Do you know?
  6. Nov 12, 2013 #5
    That makes sense. g is bounded because any infinitely differentiable function over an interval is analytic and then the following holds:http://en.wikipedia.org/wiki/Analytic_function#Alternative_characterizations (at least I think that's the reason).

    As far as the decay goes, I don't know what the ??? would. This is for a PDE course and that's all I was given as a question
  7. Nov 12, 2013 #6
    Your reason for g being bounded is correct, but it's overkill plus there is a mistake. It is enough to say g is differentiable on the interval, thus continuous, and any continuous function on a closed interval is bounded.

    If g is in ##C^ \infty## that does not mean it is analytic. The standard example is g(x) = 0 for x ≤0; g(x) =## e^{-x^2}## for x > 0. You can see that g is infinitely differentiable at x = 0. All the derivatives are 0 at x = 0. Thus its Taylor's series is identically zero, which converges fine to g for x ##\le## 0 but is nowhere near the mark for x > 0.

    For a function to be analytic at a point, its Taylor's series must converge to it within some interval around that point. For real analytic functions that is the definition.

    Re the decay, I would think that is with respect to time, but I don't see a t in the problem. ? Could you get this clarified?
  8. Nov 12, 2013 #7
    Hey, sorry for the late response but thanks for elucidating about analytic functions. As far as the decay is concerned, I can't get it clarified till tomorrow, but I really appreciated your help!
  9. Nov 12, 2013 #8
    Alright, so I think I got it. I can bound the integral with the following:


    Now, the period of sin (λx2) is 2pi/λx, which goes to zero as λ→∞. So I think the decay rate would be 1/λx?
  10. Nov 13, 2013 #9
    back to you later today
  11. Nov 13, 2013 #10
    The period of sin(## \lambda x^2##) is ##\sqrt \pi / \lambda##.

    Continuing with the problem: We have shown that ##I_λ \rightarrow 0## as ##λ \rightarrow 0##. From this point I am guessing, so don't quote me; but you can ask if this is the right kind of thing:

    By decay I think he means how fast does ##I_λ \rightarrow 0##? So we are looking at dI/dλ. That can be integrated under the integral sign to get

    ##\int_{-1}^1 g(x)x^2 cos(λx^2)dx##. Since cos(x) ≈ 1 - ##x^2##/2 the integral is approximately

    ##\int_{-1}^1 g(x)x^2 [1 -(λx^2)^2/2]dx## = ## \int_{-1}^1 g(x)x^2 dx - \frac{1}{2}λ^2\int_{-1}^1 g(x)x^6 dx ##. Both the integrals are constants with respect to λ so we can write this last expression as A -Bλ##^2##. This would be the kind of answer that is needed: we would say the decay is o(λ##^2##), where the "o" stands for order.

    Order is a measure of the general size of something. In the case of convergence we consider it to be a power of whatever variable is going to 0. The particular factors the variable may be multiplied by are irrelevant because they are constants. In this case the variable is λ; if you are defining a derivative it is usually h. When we are consider comparative bigness, we usually think of it as a power of 10 as in "your financial calculation is off by an order of magnitude".
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted