Even stronger than Sophomore's dream Sophomore's dream states that:
$$
\int_0^1x^{-x}dx=\sum_{n=1}^\infty n^{-n}
$$
and
$$
\int_0^1x^{x}dx=-\sum_{n=1}^\infty(-n)^{-n}
$$
A friend of mine noticed that numerically:
$$
\int_0^1\int_0^1(xy)^{xy}dxdy\approx  0.7834... \approx \int_0^1x^{x}dx
$$
Are these two integrals really equal?
 A: Change variable from $(x,y)$ to $(z = xy, y)$, we have
$$\begin{align}
  \int_0^1 \int_0^1 (xy)^{xy} dx dy = &\int_0^1 \int_0^y \frac{z^z}{y} dz dy
= \int_0^1 \int_z^1 \frac{z^z}{y} dy dz\\
= & \int_0^1 (- \log z) z^z dz
= \int_0^1 z^z - (1+\log z) z^z dz\\
= &\int_0^1 z^z - (z^z)' dz
= \int_0^1 z^z dz - [ z^z ]_0^1\\
= &\int_0^1 z^z dz 
\end{align}
$$
A: Yes they are equal! The trick is to use the binomial theorem to separate the integrals:
\begin{align}
\int_0^1\int_0^1(xy)^{xy}dxdy &= \int_0^1\int_0^1\exp(xy\log{xy})dxdy \\
&= \sum_{n=0}^\infty\frac{1}{n!}\int_0^1\int_0^1(xy\log{xy})^ndxdy \\
&= \sum_{n=0}^\infty\frac{1}{n!}\int_0^1\int_0^1(xy)^n(\log{x}+\log{y})^ndxdy \\
&= \sum_{n=0}^\infty\frac{1}{n!}\sum_{k=0}^nC_n^k\left(\int_0^1 x^n(\log{x})^kdx\right)\left(\int_0^1 y^n(\log{y})^{(n-k)}dy\right)
\end{align}
Then using this result:
$$
\int_0^1u^n\log{u}^m du=-m!(-\frac{1}{1+n})^{1+m}
$$
we get:
\begin{align}
\int_0^1\int_0^1(xy)^{xy}dxdy &= \sum_{n=0}^\infty\frac{1}{n!}\sum_{k=0}^nC_n^k\left(k!(-\frac{1}{1+n})^{1+k}\right)\left((n-k)!(-\frac{1}{1+n})^{1+n-k}\right) \\
 &= \sum_{n=0}^\infty\frac{1}{n!}\sum_{k=0}^nn!(-\frac{1}{1+n})^{2+n} \\
 &= \sum_{n=0}^\infty (1+n)(-\frac{1}{1+n})^{2+n} \\
 &= \sum_{n=0}^\infty (-1)^{2+n}(\frac{1}{1+n})^{1+n} \\
 &= \sum_{n=1}^\infty -(\frac{-1}{n})^{n} \\
 & = \int_0^1x^xdx
\end{align}
