Sophomore's dream: $\int_0^{1} x^{-x} \; dx = \sum_{n=1}^\infty n^{-n}$ In the solution of the so-called sophomore's dream, one of the key steps is to compute
$$\int_0^1 x^n (\log x)^n~\mathrm dx$$
using the change of variables $x = \exp\left(-\frac{u}{n+1}\right)$ to obtain the Gamma function.
This substitution to me, looks like it was pulled out of thin air. Can someone help me motivate it? How would I have thought of this substitution?
 A: Here's a bird's-eye view of the chain of arguments for how you get from the integral to the series:
$$\begin{align}
\int_{0}^{1}x^{-x}\mathrm{d}x&=\int_{0}^{1}\exp{\left(-x\log{x}\right)}\mathrm{d}x\\
&=\int_{0}^{1}\sum_{n=0}^{\infty}\frac{\left(-x\log{x}\right)^n}{n!}\mathrm{d}x\\
&=\sum_{n=0}^{\infty}\int_{0}^{1}\frac{\left(-x\log{x}\right)^n}{n!}\mathrm{d}x\\
&=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\int_{0}^{1}x^n\left(\log{x}\right)^n\mathrm{d}x\\
&...\\
&=\sum_{n=0}^{\infty}(n+1)^{-(n+1)}\\
&=\sum_{n=1}^{\infty}n^{-n}.
\end{align}$$
Thus, to to fill in the missing blanks in the middle of the proof, it will suffice to prove:
$$\int_{0}^{1}x^n\left(\log{x}\right)^n\mathrm{d}x=(-1)^n(n+1)^{-(n+1)}n!.$$
Now, even if you're not clever enough to think of the change of variables $x = \exp\left(-\frac{u}{n+1}\right)$ to obtain the Gamma function, you can still figure out this integral by integrating by parts iteratively, as follows:
$$\begin{align}
\int_{0}^{1}x^n\left(\log{x}\right)^n\mathrm{d}x&=\frac{1}{n+1}x^{n+1}\left(\log{x}\right)^n\bigg{|}_{0}^{1}-\int_{0}^{1}\frac{1}{n+1}x^{n+1}\frac{n\left(\log{x}\right)^{n-1}}{x}\mathrm{d}x\\
&=-\frac{n}{n+1}\int_{0}^{1}x^{n}\left(\log{x}\right)^{n-1}\mathrm{d}x\\
&=(-1)^2\frac{n(n-1)}{(n+1)^2}\int_{0}^{1}x^{n}\left(\log{x}\right)^{n-2}\mathrm{d}x\\
&=(-1)^3\frac{n(n-1)(n-2)}{(n+1)^3}\int_{0}^{1}x^{n}\left(\log{x}\right)^{n-3}\mathrm{d}x\\
&...\\
&=(-1)^n\frac{n!}{(n+1)^n}\int_{0}^{1}x^{n}\left(\log{x}\right)^{n-n}\mathrm{d}x\\
&=(-1)^n\frac{n!}{(n+1)^{n+1}}.
\end{align}$$
However, if you know that $n!=\Gamma(n+1)=\int_{0}^{\infty}u^ne^{-u}\mathrm{d}u$, you can see that the identity we are trying to prove is:
$$\begin{align}
\int_{0}^{1}x^n\left(\log{x}\right)^n\mathrm{d}x&=(-1)^n(n+1)^{-(n+1)}\int_{0}^{\infty}u^ne^{-u}\mathrm{d}u\\
&=(n+1)^{-1}\int_{0}^{\infty}\frac{(-1)^nu^n}{(n+1)^n}e^{-u}\mathrm{d}u\\
&=(n+1)^{-1}\int_{0}^{\infty}\left(\frac{-u}{(n+1)}\right)^ne^{-u}\mathrm{d}u
\end{align}$$
This last line strongly suggests the substitution,
$$\log{x}=\frac{-u}{(n+1)},$$
or evquivalently,
$$x=\exp{\left(\frac{-u}{n+1}\right)}.$$
