$$\text{Compute} :\iint_{[0,1]^2} (xy)^{xy} dxdy$$

I am thinking about changing the variable, $x=u,y={v \over u}$.But it doesn't work.

I just found that the answer is$\int_0^1 t^t dt$.Maybe my idea is right?

  • $\begingroup$ Just to be clear, is this $(xy)^{xy}$ or $x\, y^{xy}$? $\endgroup$ Oct 21, 2014 at 15:54
  • $\begingroup$ @GrumpyParsnip My bad,i'll correct it. $\endgroup$ Oct 21, 2014 at 15:55
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    $\begingroup$ In response to your edit, if you set $t= xy$ and $s=y$, the Jacobian determinant is $\partial(s,t)/\partial(x,y)=y$, implying $\partial(x,y)/\partial(s,t)=1/y=1/s$. So you get $\int_0^1\int_0^s t^t (1/s)\,dt\,ds,$ but I don't see how to get to $\int_0^1 t^t\,dt$. $\endgroup$ Oct 21, 2014 at 16:09

1 Answer 1


Building on Grumpy's comment, changing the order of integration in $\int_0^1 \int_0^s t^t (1/s) \ dt\ ds$ gives $$\int_0^1 \int_t^1 t^t (1/s)\ ds\ dt = \int_0^1 t^t (-\ln t)\ dt.$$

The claim in the question thus amounts to showing that $\int_0^1 t^t (1+\ln t)\ dt = 0$, which follows readily from the substitution $u = t \ln t$.

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    $\begingroup$ It is interesting to note that in this case you can switch the order of integration because $t^t 1/s$ is non-negative ; this is Fubini's theorem. +1 $\endgroup$ Oct 21, 2014 at 16:24
  • $\begingroup$ Nice. I got that expression, didn't see what to do, and posted it as a comment. $\endgroup$ Oct 21, 2014 at 16:27
  • $\begingroup$ Nice!Thanks a lot! $\endgroup$ Oct 21, 2014 at 16:37
  • $\begingroup$ @PatrickDaSilva I'm sorry that I can't figure out why $\int_0^1 t^t (1+\ln t)\ dt = 0$ by the substitution $u = t \ln t$.Could you possibly explain it in a little more detail? $\endgroup$ Oct 21, 2014 at 16:49
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    $\begingroup$ @GEE20151011 : Next time ask the answerer of the question for that haha :) $e^u = e^{t \ln t} = (e^{\ln t})^t = t^t$ and $du = (\ln t + 1) dt$. So you get the integral $\int_{u(0)}^{u(1)} e^u du = 0$ (because $u(0) = \lim_{t \to 0^+} t \ln t = 0$ and $u(1) = 1 \ln 1 = 0$). $\endgroup$ Oct 21, 2014 at 20:04

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