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Math Help - Unknown limits

  1. #1
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    Jun 2009
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    8

    Unknown limits

    Can you solve this two limits (with demostration, obviously)?

    Ei(n) is the exponential integral function
    Attached Thumbnails Attached Thumbnails Unknown limits-image001.png   Unknown limits-image003.png   Unknown limits-image005.png  
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  2. #2
    Super Member
    Joined
    Aug 2008
    Posts
    903
    I'd like to suggest a start but not sure it's sound. Perhaps others can comment.

    First, clearly define the integral:

    \text{Ei}_n(x)=\int_1^{\infty}\frac{e^{xt}}{t^n}dt

    so need to consider:

    \sum_{n=1}^{\infty}\int_1^{\infty}\frac{e^{xt}}{t^  n}dt-\sum_{n=1}^{\infty}\frac{e^{-x}}{n+x}

    Now express the last expression as the same type integral:

    \frac{e^{-x}}{n+x}=\int_1^{\infty}e^{1-t}\frac{e^{-x}}{n+x}dt

    Combining both sums, I obtain:

    \sum_{n=0}^{\infty}\int_1^{\infty}\left( \frac{e^{-xt}}{t^n}-\frac{e^{-x}e^{1-t}}{n+x}\right)dt

    Now, at this point, how about using the Integral Test? I'm however not sure that's valid and I'd need to work on that part. If it were valid, I could then write:

    \int_{1}^{\infty}\int_1^{\infty} \frac{e^{-xt}}{t^n}-\frac{e^{-x}e^{1-t}}{n+x}dtdn

    Switching the order:

    \int_{1}^{\infty}\int_1^{\infty} \frac{e^{-xt}}{t^n}-\frac{e^{-x}e^{1-t}}{n+x}dndt

    I get:

    \int_{1}^{\infty}\left\{\left(-\frac{1}{\ln(t)}e^{-n\ln(t)}\Biggr|_1^{\infty}\right)-e^{-x}e^{1-t}\ln(n+x)\Biggr|_1^{\infty}\right\}dt

    and that last term diverges. So if the steps I've taken so far are valid, and I could apply the integral test in this fashion, then I would conclude the original sum diverges.

    . . . Gee whiz, you guys are way better at this than me but I'm big at tries.
    Last edited by mr fantastic; July 10th 2011 at 04:26 PM.
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