Integrating $\int_0^1 \frac{\ln x}{x^x}\, dx.$ I am trying to integrate this.
$$
\int_0^1 \frac{\ln x}{x^x}\, dx.
$$
Thanks.
I thought
$$
\int_0^1 \ln x \, x^{-x} \, dx=\int_0^1 \ln x\, e^{-x\ln x}\, dx=\sum_{n=0}^\infty\frac{(-1)^n}{n!}\int_0^1 x^n \ln x (\ln x)^n\, dx=\\\sum_{n=0}^\infty\frac{(-1)^n}{n!}\int_0^1 x^n (\ln x)^{n+1}\,dx
$$
but now I am confused because I cannot solve this integral.  I do know that
$$
\int_0^1 x^{-x}\, dx= \sum_{n=1}^\infty n^{-n}.
$$
 A: Since $(x\ln x)'=\ln x+1$, we have
$$x^{-x}(\ln x+1)=-(e^{-x\ln x})'.$$
Integrating both sides of this identity between $0$ and $1$ and using that
$$\lim_{x\rightarrow 0^+}x\ln x= \lim_{x\rightarrow1}x\ln x=0,$$
one finds that the integral we are looking for is equal to minus the Sophomore's dream constant:
$$\int_0^1x^{-x}\ln x\,dx=-\int_0^1x^{-x}dx=-\sum_{n=1}^{\infty}n^{-n}.$$
No closed expression for this constant is known so far.
A: As a brute force verification of O.L.'s very clever observation, and continuing the approach started in the question,
$$
\begin{align}
\int_0^1\log(x)x^{-x}\,\mathrm{d}x
&=\int_0^1\log(x)e^{-x\log(x)}\,\mathrm{d}x\\
&=\sum_{k=0}^\infty\frac1{k!}\int_0^1x^k\log(x)^{k+1}\,\mathrm{d}x\\
&=\sum_{k=0}^\infty\frac1{k!}\int_0^\infty e^{-(k+1)t}(-t)^{k+1}\,\mathrm{d}t\\
&=\sum_{k=0}^\infty\frac{(-1)^{k+1}}{(k+1)^{k+2}k!}\int_0^\infty e^{-u}\,u^{k+1}\,\mathrm{d}u\\
&=\sum_{k=0}^\infty\frac{(-1)^{k+1}}{(k+1)^{k+2}k!}(k+1)!\\
&=\sum_{k=0}^\infty\frac{(-1)^{k+1}}{(k+1)^{k+1}}\\
\end{align}
$$
