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?

Change variable from $(x,y)$ to $(z = xy, y)$, we have
$$\begin{align}
  \int_0^1 \int_0^1 (xy)^{xy} dx dy = &\int_0^1 \int_0^y \frac{z^z}{y} dz dy
= \int_0^1 \int_z^1 \frac{z^z}{y} dy dz\\
= & \int_0^1 (- \log z) z^z dz
= \int_0^1 z^z - (1+\log z) z^z dz\\
= &\int_0^1 z^z - (z^z)' dz
= \int_0^1 z^z dz - [ z^z ]_0^1\\
= &\int_0^1 z^z dz 
\end{align}
$$
