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Math Help - Evaluating this integral

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    Evaluating this integral

    {\int^\infty_1} \frac{lnx}{x^2}dx

    I am not sure where to start. I tried u substitution, but it didn't seem to get me anywhere. This is to evaluate an infinite series using the integral test. I could use a hint. TIA.
    Last edited by wondering; July 28th 2013 at 05:49 PM. Reason: I forgot the dx
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    Re: Evaluating this integral

    Quote Originally Posted by wondering View Post
    {\int^\infty_1} \frac{lnx}{x^2}dx
    I am not sure where to start. I tried u substitution, but it didn't seem to get me anywhere. This is to evaluate an infinite series using the integral test. I could use a hint. TIA.
    Have a look at this.
    Thanks from wondering
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    Re: Evaluating this integral

    Is integration by parts the only way. The solution manual doesn't show any steps, but they will usually mention if they have used by parts, and they don't. Thank you for the link.
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    Re: Evaluating this integral

    Quote Originally Posted by wondering View Post
    Is integration by parts the only way. The solution manual doesn't show any steps, but they will usually mention if they have used by parts, and they don't. Thank you for the link.
    On the page I posted there is a "show steps" option. Click it and see the complete solution.
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    Re: Evaluating this integral

    Quote Originally Posted by wondering View Post
    {\int^\infty_1} \frac{lnx}{x^2}dx

    I am not sure where to start. I tried u substitution, but it didn't seem to get me anywhere. This is to evaluate an infinite series using the integral test. I could use a hint. TIA.
    You say you "tried u substitution" but don't say what substitution. Was it "u= ln(x)"? That would seem the obvious substitution. Of course then du= (1/x)dx so the integral becomes \int_1^\infty \frac{ln(x)}{x}(1/x)dx= \int_0^\infty \frac{u}{x}du.

    And are you saying "it didn't seem to get me anywhere" because of that "x" still in the integrand? Well, if u= ln(x), then x= e^u so the integral becomes
    \int_0^\infty \frac{u}{e^u}du= \int_0^\infty ue^{-u}du which can be integrated "by parts".
    Thanks from wondering
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    Re: Evaluating this integral

    Quote Originally Posted by HallsofIvy View Post
    You say you "tried u substitution" but don't say what substitution. Was it "u= ln(x)"? That would seem the obvious substitution. Of course then du= (1/x)dx so the integral becomes \int_1^\infty \frac{ln(x)}{x}(1/x)dx= \int_0^\infty \frac{u}{x}du.

    And are you saying "it didn't seem to get me anywhere" because of that "x" still in the integrand? Well, if u= ln(x), then x= e^u so the integral becomes
    \int_0^\infty \frac{u}{e^u}du= \int_0^\infty ue^{-u}du which can be integrated "by parts".
    Yes, it was the x in the denominator that I didn't know what to do with. Thanks for the reply, it makes sense now. I should have written out all of my steps, but it takes me forever to get the latex right.
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