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Thread: Integral limit

  1. #1
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    Integral limit

    Does the integral in the attached image converge?

    Can you give me a hint how to start solving this question. Thanks
    Attached Thumbnails Attached Thumbnails Integral limit-5c.jpg  
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  2. #2
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    Re: Integral limit

    Here's the informal argument. I'll leave it to you to make it rigorous.

    The singularity occurs at t=1, so we need to think about what the function $\displaystyle (\log|\log t|)^7$ looks like when t is close to 1.

    For t close to 1, $\displaystyle |\log t|$ is approximately 1t, so we should look at the function $\displaystyle \bigl(\log(1-t)\bigr)^7.$ In fact, that function is negative for 0<t<1, so it would be more convenient to look at its negative, namely $\displaystyle \bigl|\log(1-t)\bigr|^7.$

    As x decreases to 0, |log x| increases to infinity, but more slowly than any negative power of x: $\displaystyle |\log x|\leqslant x^{-\alpha}$ for any $\displaystyle \alpha>0.$ Put x = 1t to get $\displaystyle \bigl|\log(1-t)\bigr|^7 \leqslant (1-t)^{-7\alpha}.$ Choosing $\displaystyle \alpha = 1/14$, you see that $\displaystyle \bigl|\log(1-t)\bigr|^7 \leqslant (1-t)^{-1/2}.$ The integral $\displaystyle \int_{1/2}^1(1-t)^{-1/2}dt$ converges, hence (by the comparison test for improper integrals) so should the integral that you started with.
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  3. #3
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    Re: Integral limit

    Quote Originally Posted by Opalg View Post
    For t close to 1, $\displaystyle |\log t|$ is approximately 1t
    How do I know this is true?

    Quote Originally Posted by Opalg View Post
    As x decreases to 0, |log x| increases to infinity, but more slowly than any negative power of x: $\displaystyle |\log x|\leqslant x^{-\alpha}$ for any $\displaystyle \alpha>0.$
    How do I prove this?

    Thanks
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: Integral limit

    Verify $\displaystyle \lim_{t\to 1^-}\frac{|\log t|}{1-t}=\ldots=1$ and $\displaystyle \lim_{x\to 0^+}\frac{|\log x|}{x^{-\alpha}}=\ldots=0\;\;(\alpha>0)$
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  5. #5
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    Re: Integral limit

    Quote Originally Posted by FernandoRevilla View Post
    Verify $\displaystyle \lim_{t\to 1^-}\frac{|\log t|}{1-t}=\ldots=1$
    Thank you. I managed to prove the second limit, but I don't know how to prove this one.

    Quote Originally Posted by Opalg View Post
    The integral $\displaystyle \int_{1/2}^1(1-t)^{-1/2}dt$ converges
    And I how do I prove this?
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Re: Integral limit

    Use the substitution $\displaystyle u=1-t$ and the well known property: $\displaystyle \int_{0}^1\frac{du}{u^p}$ converges iff $\displaystyle p<1$ .
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