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Thread: 2 calculus quesions

  1. #1
    Dec 2009

    2 calculus quesions

    Hey there,

    I'll be glad to get some help in the following questions:

    1. Let g(x) be a differentiable function at $\displaystyle (0,1]$ which satisfies: $\displaystyle x^{2}g(x) $ is a monotonic ascending function at
    $\displaystyle (0,1]$ and: $\displaystyle lim_{x \to 0^{+}} x^{2}g(x)=0 $, g'(x) is continous at $\displaystyle (0,1]$.
    Check whether the integral $\displaystyle \int_{0}^{1} g(x)sin(\frac{1}{x})dx $ converges.

    2. Let f(x) be a function defined by: $\displaystyle f(x)=\Sigma_{n=1}{\infty}a_{n}sin(nx) $ .
    Prove that if the series $\displaystyle \Sigma_{n=1}^{\infty}n|a_{n}| $ converges then f(x) is differentiable for every real x.

    Thanks in advance
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  2. #2
    MHF Contributor chisigma's Avatar
    Mar 2009
    near Piacenza (Italy)
    Setting $\displaystyle \frac{1}{x}=t$ the integral becomes...

    $\displaystyle \int_{0}^{1} g(x)\cdot \sin \frac{1}{x}\cdot dx = \int_{1}^{\infty} g(\frac{1}{t})\cdot \frac{\sin t}{t^{2}}\cdot dt $ (1)

    Now the function...

    $\displaystyle \gamma (t) = \frac{g(\frac{1}{t})}{t^{2}}$ (2)

    ... is monotonically decreasing for $\displaystyle t>1$ nd ...

    $\displaystyle \lim_{t \rightarrow \infty} \frac{g(\frac{1}{t})}{t^{2}}=0$ (3)

    ... so that is...

    $\displaystyle |\int_{1}^{\infty} g(\frac{1}{t})\cdot \frac{\sin t}{t^{2}}\cdot dt| < |\int_{1}^{\pi} \gamma(t)\cdot \sin t\cdot dt| + \pi \cdot |\sum_{n=1}^{\infty} (-1)^{n} \gamma (n\cdot \pi)| $ (4)

    ... and the integral converges...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Last edited by chisigma; Apr 18th 2010 at 05:15 AM. Reason: Added a 'pi' in (4)...
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  3. #3
    Dec 2009
    Thanks a lot
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