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Math Help - help with this integral problem

  1. #1
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    help with this integral problem

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  2. #2
    MHF Contributor kalagota's Avatar
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    for the first one, it just follows directly from m \leq f(x) \leq M by taking the definite integrals in all parts (since integral preserves the direction of the inequality). Also note that constants can go out of the integral sign.

    for the second part, since g(x) \geq 0, the inequality does not change i.e. mg(x) \leq f(x)g(x) \leq Mg(x) and you can take the integrals.

    the next two are applications.. you can do it.
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  3. #3
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    Quote Originally Posted by szpengchao View Post
    First apply parts to get,

    \int_{0}^{\frac1{\sqrt{n}}} nf(x) e^{-nx} \, dx = -f(x) e^{-nx} \bigg{|}_{0}^{\frac1{\sqrt{n}}} + \int_{0}^{\frac1{\sqrt{n}}} f'(x) e^{-nx} \, dx

     \lim_{n \to  \infty} -f(x) e^{-nx} \bigg{|}_{0}^{\frac1{\sqrt{n}}}

     = \lim_{n \to  \infty} -f(\frac1{\sqrt{n}}) e^{-\sqrt{n} x} + f(0) = f(0)

    [That is because \lim_{n \to  \infty} -f(\frac1{\sqrt n})e^{-\sqrt{n}x} = 0 since e^{-\sqrt{n}x} \to 0 as n \to \infty]

    We claim \int_{0}^{\frac1{\sqrt{n}}} f'(x) e^{-nx} \, dx = 0

    First observe that 0 < e^{-nx} < 1

    So 0 < \int_{0}^{\frac1{\sqrt{n}}}  e^{-nx}f'(x)\, dx < \int_{0}^{\frac1{\sqrt{n}}}  f'(x) \, dx

    0 < \lim_{n \to \infty} \int_{0}^{\frac1{\sqrt{n}}}  e^{-nx}f'(x)\, dx < \lim_{n \to \infty} f\left(\frac1{\sqrt{n}}\right)  - f(0) = f(0)\text{(Why?) }- f(0) = 0

    Thus claim proved by sandwich theorem...

    So
    \lim_{n \to  \infty} \int_{0}^{\frac1{\sqrt{n}}} nf(x) e^{-nx} \, dx  = f(0)


    EDIT: I realised this halfway through... My proof is incomplete... I assumed (1) f is differentiable (2) f'(x) > 0.... Let me think about it a little more....
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  4. #4
    MHF Contributor kalagota's Avatar
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    as i have said, the questions are just applications of the previous.

    note that, f is continuous in a closed and bounded interval, then you can use Maximum-Minimum theorem, that is f is also bounded. Also, f is integrable.

    Let m, M be real numbers such that m \leq f(x) \leq M for x \in [0,1]. Applying the first and second facts, we have m \int_0^{1/\sqrt{n}} n e^{-nx}dx \leq \int_0^{1/\sqrt{n}} nf(x)e^{-nx}dx \leq M \int_0^{1/\sqrt{n}} n e^{-nx}dx (since 1/\sqrt{n} \in [0,1])

    i just wonder if it equal f(0) or 0 unless f(0) = 0
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  5. #5
    Lord of certain Rings
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    Quote Originally Posted by kalagota View Post
    as i have said, the questions are just applications of the previous.

    note that, f is continuous in a closed and bounded interval, then you can use Maximum-Minimum theorem, that is f is also bounded. Also, f is integrable.

    Let m, M be real numbers such that m \leq f(x) \leq M for x \in [0,1]. Applying the first and second facts, we have m \int_0^{1/\sqrt{n}} n e^{-nx}dx \leq \int_0^{1/\sqrt{n}} nf(x)e^{-nx}dx \leq M \int_0^{1/\sqrt{n}} n e^{-nx}dx (since 1/\sqrt{n} \in [0,1])

    i just wonder if it equal f(0) or 0 unless f(0) = 0
    That does not lead us anywhere, but tell us that the integral is between M and m!
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