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Math Help - Double integral

  1. #1
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    Double integral

    Evaluate

    \int_0^1\int_0^1\frac{dx\,dy}{\ln^{3/2}\left(\dfrac1{xy}\right)}
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  2. #2
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    For the inner integral, substitute u=xy,

    \int_0^1 {\int_0^1 {\frac{{dx\,dy}}<br />
{{\Big[ { - \ln (xy)} \Big]^{3/2} }}} } = \int_0^1 {\int_0^y {\frac{{du\,dy}}<br />
{{y\left( { - \ln u} \right)^{3/2} }}} } = \int_0^1 {\int_u^1 {\frac{{dy\,du}}<br />
{{y\left( { - \ln u} \right)^{3/2} }}} } .

    So

    \int_0^1 {\int_0^1 {\frac{{dx\,dy}}<br />
{{\Big[ { - \ln (xy)} \Big]^{3/2} }}} } = \int_0^1 {\frac{{ - \ln u}}<br />
{{\left( { - \ln u} \right)^{3/2} }}\,du} = \int_0^1 {\frac{1}<br />
{{\sqrt { - \ln u} }}\,du} = \sqrt \pi .
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  3. #3
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    i don't think it works because that function is not defined for all x and y in [0,1].

    for exaomple when x=y=1 we have

    \frac{1}{(\log{1})^{\frac{3}{2}}}

     = \frac{1}{0}
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  4. #4
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    It is okay to do this. It is an improper integral.
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