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Math Help - Uniform Convergence of Inegral

  1. #1
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    Uniform Convergence of Inegral

    Prove that the integral \int_{0}^{1}\frac{\sin\alpha x}{\sqrt{|x-\alpha|}}\,dx converges uniformly for \alpha\in[0,1].

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    Re: Uniform Convergence of Inegral

    I'm not sure what "converges uniformly" means in this situation. Absolute convergence, as in the integral of the absolute value of the function converges? If so, then it seems the answer is yes.

    With the exception of \alpha \in \{0, 1\}, you split the integral up into two integrals having its "bad" point as an endpoint, and then each of those integrals converges in absolute value. Write:

    \int_{0}^{1} \frac{\sin(\alpha x)}{\sqrt{|\alpha - x|}} dx = \int_{0}^{\alpha} \frac{\sin(\alpha x)}{\sqrt{\alpha - x}} dx + \int_{\alpha}^{1} \frac{\sin(\alpha x)}{\sqrt{x - \alpha }} dx ,

    and then substitute u = 1 - \frac{x}{\alpha} for the first, and u = \frac{x}{\alpha} - 1 for the second. Then simplify and use the trig angle sum formulas.

    What you get is that, for each \alpha \neq 0, after these standard integral manipulations, you're left asking if

    \int_{0}^{whatever} \frac{|\sin(u)|}{\sqrt{u}} du and \int_{0}^{whatever} \frac{|\cos(u)|}{\sqrt{u}} du converge. They do.
    Last edited by johnsomeone; September 13th 2012 at 04:00 PM.
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    Re: Uniform Convergence of Inegral

    If that felt "too Riemannian", consider the function g(x) = \frac{1}{\sqrt{-x}} for -1 \le x < 0, g(0) = 1, and \frac{1}{\sqrt{x}} for 0 < x \le 1.

    Then g \in L^1([-1, 1]), so h(x) = g(x-\alpha) \in L^1([-1+\alpha, 1+\alpha]), so \tilde{h} = h|[0, 1] \in L^1([0, 1]).

    Then on x \in [0,1], \ \frac{\sin(\alpha x)}{\sqrt{|\alpha-x|}} = \sin(\alpha x)g(x - \alpha) = \sin(\alpha x)\tilde{h}(x), so

    \int_{0}^{1}| \ \frac{\sin(\alpha x)}{\sqrt{|\alpha-x|}}\ | \ dx = \int_{0}^{1}| \ \sin(\alpha x)\tilde{h}(x) \ | \ dx  \le \int_{0}^{1}|\tilde{h}(x)| dx < \infty.
    Last edited by johnsomeone; September 14th 2012 at 02:42 PM.
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