Double integral

**liyi** · November 27th 2007, 01:39 PM

Evaluate

$\int_0^1\int_0^1\frac{dx\,dy}{\ln^{3/2}\left(\dfrac1{xy}\right)}$

**Krizalid** · November 27th 2007, 03:18 PM

For the inner integral, substitute $u=xy,$

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

So

$\int_0^1 {\int_0^1 {\frac{{dx\,dy}} {{\Big[ { - \ln (xy)} \Big]^{3/2} }}} } = \int_0^1 {\frac{{ - \ln u}} {{\left( { - \ln u} \right)^{3/2} }}\,du} = \int_0^1 {\frac{1} {{\sqrt { - \ln u} }}\,du} = \sqrt \pi .$

**Aradesh** · November 27th 2007, 03:19 PM

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

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

$= \frac{1}{0}$

**ThePerfectHacker** · November 27th 2007, 07:06 PM

It is okay to do this. It is an improper integral.

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