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Math Help - Stuck on Another Direct Comparison Integral

  1. #1
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    Stuck on Another Direct Comparison Integral



    I don't seem to be able to do these very well.

    Use the Comparison Theorem to determine whether the integral is convergent or divergent:

    \int_0^1 \frac{e^{-x}}{\sqrt{x}}dx

    I see two functions that might be of assistance:

    on [0,1], e^{-x} \leq 1 so \frac{e^{-x}}{\sqrt{x}} \leq \frac{1}{\sqrt{x}}

    But that won't help because \int_0^1 \frac{1}{\sqrt{x}}dx is divergent, so it says nothing about a function that is less than it.


    Then, I looked at the fact that on [0,1] \sqrt{x} \leq 1 so then \frac{e^{-x}}{\sqrt{x}} \geq e^{-x}. But \int e^{-x} is a proper integral, so it says nothing about an integral that is bigger than it.

    AAAAAGH. I can't seem to get any of these figure out. What am I doing wrong?
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  2. #2
    Grand Panjandrum
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    Re: Stuck on Another Direct Comparison Integral

    Quote Originally Posted by joatmon View Post


    I don't seem to be able to do these very well.

    Use the Comparison Theorem to determine whether the integral is convergent or divergent:

    \int_0^1 \frac{e^{-x}}{\sqrt{x}}dx

    I see two functions that might be of assistance:

    on [0,1], e^{-x} \leq 1 so \frac{e^{-x}}{\sqrt{x}} \leq \frac{1}{\sqrt{x}}

    But that won't help because \int_0^1 \frac{1}{\sqrt{x}}dx is divergent, so it says nothing about a function that is less than it.


    Then, I looked at the fact that on [0,1] \sqrt{x} \leq 1 so then \frac{e^{-x}}{\sqrt{x}} \geq e^{-x}. But \int e^{-x} is a proper integral, so it says nothing about an integral that is bigger than it.

    AAAAAGH. I can't seem to get any of these figure out. What am I doing wrong?
    for x \in [0,1] we have 1 \ge e^{-x}\ge e^{-1}

    Therefore for \varepsilon \in (0,1]:

    \int_{\varepsilon}^1 \frac{1}{\sqrt{x}}dx\ge \int_{\varepsilon}^1 \frac{e^{-x}}{\sqrt{x}}dx \ge \int_{\varepsilon}^1 \frac{e^{-1}}{\sqrt{x}}dx

    Hence the integral converges/diverges as \int_{0}^1 \frac{1}{\sqrt{x}}dx converges/diverges

    CB
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