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Math Help - existence of the definite integral

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    existence of the definite integral

    I do not know how to proof that integral \int^{\infty}_{1} \frac{(lnx)^{\pi}}{(x-1)^3} dx exists. I have used Taylor or convergence tests but I do not see the next steps.
    Last edited by Camille91; August 30th 2011 at 03:27 AM.
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    Re: existence of the definite integral

    Quote Originally Posted by Camille91 View Post
    I do not know how to proof that integral \int^{\infty}_{1} \frac{(lnx)^{\pi}}{(x-1)^3} dx exists. I have used Taylor or convergence tests but I do not see the next steps.
    Use the substitution x-1=\xi and then observe that \lim_{\xi \rightarrow 0}\frac{\ln ^{\pi} (1+\xi)}{\xi^{3}}=0...

    Kind regards

    \chi \sigma
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    Re: existence of the definite integral

    I guess now I should do:  \int^{\infty}_{1} \frac{(lnx)^{\pi}}{(x-1)^3} dx =\underbrace{ \int^{A}_{1} \frac{(lnx)^{\pi}}{(x-1)^3} dx}_{1} +\underbrace{\int^{\infty}_{A} \frac{(lnx)^{\pi}}{(x-1)^3} dx}_{2} for A>1. With Your observation I conclude the first integral converges but how about the second one?

    Thank you for your last post.
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    MHF Contributor chisigma's Avatar
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    Re: existence of the definite integral

    Quote Originally Posted by Camille91 View Post
    I guess now I should do:  \int^{\infty}_{1} \frac{(lnx)^{\pi}}{(x-1)^3} dx =\underbrace{ \int^{A}_{1} \frac{(lnx)^{\pi}}{(x-1)^3} dx}_{1} +\underbrace{\int^{\infty}_{A} \frac{(lnx)^{\pi}}{(x-1)^3} dx}_{2} for A>1. With Your observation I conclude the first integral converges but how about the second one?

    Thank you for your last post.
    The second integral converges because is...

    \lim_{x \rightarrow \infty} \frac{\ln ^{a} x}{x^{b}} = 0\ ,\ a>0\ , b>0 (1)

    ... no matter what are a and b...

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    \chi \sigma
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    Re: existence of the definite integral

    Using  \lim_{x \rightarrow \infty} \frac{\ln ^{\pi}x}{x^3}=0 I can prove the integer satisfied necessary condition but it is not sufficient. On the other hand in convergence tests limes to zero say nothing (I mean limit comparison test). Maybe I am wrong?
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    MHF Contributor chisigma's Avatar
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    Re: existence of the definite integral

    Quote Originally Posted by Camille91 View Post
    Using  \lim_{x \rightarrow \infty} \frac{\ln ^{\pi}x}{x^3}=0 I can prove the integer satisfied necessary condition but it is not sufficient. On the other hand in convergence tests limes to zero say nothing (I mean limit comparison test). Maybe I am wrong?
    What I intended to say is that for x 'large enough' is...

    \frac{\ln^{\pi} x}{x^{3}}< \frac{1}{x^{3-\varepsilon}} (1)

    ... no matter 'how small' is \varepsilon>0. Now the integral...

    \int_{a}^{\infty}\frac{dx}{x^{\alpha}}\ ,\ a>0 (2)

    .... converges for all \alpha>1, so that also Your integral converges...

    Kind regards

    \chi \sigma
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  7. #7
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    Re: existence of the definite integral

    You are absolutely right and it solved the whole problem. Thank you.
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