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Math Help - Caculation a limit ?

  1. #1
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    Caculation a limit ?

    How do I show \displaystyle\lim_{x\rightarrow 1^+ }\int_{x^2}^{x^4}\!\frac{1}{\ln t}\, dt=\ln 2 ?
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  2. #2
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    Quote Originally Posted by bigli View Post
    How do I show \displaystyle\lim_{x\rightarrow 1^+ }\int_{x^2}^{x^4}\!\frac{1}{\ln t}\, dt=\ln 2 ?

    If you've already done power series and stuff, we get (Bronshtein-Semendiaev, "Manuel of Mathetmatics") that

    \int\frac{dt}{\ln t} =\ln\ln t+\ln t+\sum^\infty_{n=2}\frac{(\ln t)^n}{n\cdot n!}

    Use this in your definite integral and pass to the limit and you'll get your answer.

    Tonio
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  3. #3
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    It doesn't work.Is there any easier way?
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  4. #4
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    Quote Originally Posted by bigli View Post
    It doesn't work.Is there any easier way?


    Oh, it works and very fine , thank you! Perhaps there's an easier way but I can't see it.

    Tonio
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  5. #5
    Super Member Random Variable's Avatar
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    According to Maple, the series expansion of  \frac{1}{\ln t} about  t = 1 is  \frac{1}{t-1} + \frac{1}{2} - \frac{1}{12} (t-1) + \frac{1}{24} (t-1)^{2} - \frac{19}{720} (t-1)^{3} + ... (EDIT: You can derive it by starting with the Taylor series for  \ln t centered at  t=1 and using long division.)

    Since we're integrating over values close to 1,  \frac{1}{\ln t} is approximately  \frac{1}{t-1}.

    so  \lim_{x \to 1^{+}} \int^{x^{4}}_{x^{2}} \frac{1}{\ln t} \ dt = \lim_{x \to 1^{+}} \int^{x^{4}}_{x^{2}} \frac{1}{t-1} \ dt

     = \lim_{x \to 1^{+}}  \ln (t-1)  \Big|^{x^{4}}_{x^{2}}

     =\lim_{x \to 1^{+}} \Big( \ln (x^{4}-1) - \ln (x^{2}-1) \Big)

     = \lim_{x \to 1^{+}}\ln \Big(\frac{x^{4}-1}{x^{2}-1}\Big)

     = \lim_{x \to 1^{+}} \ln (x^{2}+1)  = \ln 2
    Last edited by Random Variable; June 2nd 2010 at 01:46 PM.
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    Quote Originally Posted by Random Variable View Post
    According to Maple, the series expansion of  \frac{1}{\ln t} about  t = 1 is  \frac{1}{t-1} + \frac{1}{2} - \frac{1}{12} (t-1) + \frac{1}{24} (t-1)^{2} - \frac{19}{720} (t-1)^{3} + ... (EDIT: You can derive it by starting with the Taylor series for  \ln t centered at  t=1 and using long division.)

    Since we're integrating over values close to 1,  \frac{1}{\ln t} is approximately  \frac{1}{t-1}.


    And what happened to +\frac{1}{2} above...?

    Tonio


    so  \lim_{x \to 1^{+}} \int^{x^{4}}_{x^{2}} \frac{1}{\ln t} \ dt = \lim_{x \to 1^{+}} \int^{x^{4}}_{x^{2}} \frac{1}{t-1} \ dt

     = \lim_{x \to 1^{+}} \ln (t-1) \Big|^{x^{4}}_{x^{2}}

     =\lim_{x \to 1^{+}} \Big( \ln (x^{4}-1) - \ln (x^{2}-1) \Big)

     = \lim_{x \to 1^{+}}\ln \Big(\frac{x^{4}-1}{x^{2}-1}\Big)

     = \lim_{x \to 1^{+}} \ln (x^{2}+1) = \ln 2
    .
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  7. #7
    Super Member Random Variable's Avatar
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    As t approaches 1 from the right,  \frac{1}{t-1} is much greater than  \frac{1}{2} . You can include it if you want. You'll still get the same answer.
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  8. #8
    MHF Contributor Drexel28's Avatar
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    It is fairly easy to verify that for t>0 we have that \frac{t-1}{t}\leqslant\ln(t)\leqslant t-1 and so \frac{1}{t-1}\leqslant\frac{1}{\ln(t)}\leqslant\frac{t}{t-1}. It follows that \int_{x^2}^{x^4}\frac{dt}{t-1}\leqslant \int_{x^2}^{x^4}\frac{dt}{\ln(t)}\leqslant\int_{x^  2}^{x^4}\frac{t}{t-1}dt. Note though that the LHS is just \ln\left(x^4-1\right)-\ln(x^2-1)=\ln(x^2+1)\overset{x\to 1^+}{\longrightarrow}\ln(2) and  \int_{x^2}^{x^4}\frac{t}{t-1}dt=\left(x^4-x^2\right)+\left(\ln(x^4-1)-\ln(x^2-1)\right)\overset{x\to 1}{\longrightarrow}\ln(2). The conclusion follows from the Squeeze Theorem.
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  9. #9
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    Thanks DREXEL28. Every thing is OK!. But only, your inequalities should be strictly and they are correct for t>1 .
    Last edited by bigli; June 5th 2010 at 08:26 PM.
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