prove that $\displaystyle\int_0^1 \dfrac{1}{x^x}dx = 1 + \dfrac{1}{2^2} + \dfrac{1}{3^3} + \dfrac{1}{4^4} + \ldots $ $\displaystyle\int_0^1 \dfrac{1}{x^x}dx = 1 + \dfrac{1}{2^2} + \dfrac{1}{3^3} + \dfrac{1}{4^4} + \ldots $
the only idea I have is using the series expansion of $x^{-x} \approx 1 - x\log x + \dfrac{(x\log x)^2}{2!} - \ldots$. But it ends up little complicated, any idea?
 A: Yes go ahead
$$S=\int_{0}^{1} x^{-x}  dx= \int_{0}^{1} e^{-x\ln x} dx =\int_{0}^{1} \sum_{k=0}^{\infty} \frac{(-x \ln x)^k}{k!}dx.$$
Let $x=e^{-t}$,and use $\int_{0}^{\infty} t^n e^{-at}dt=\frac{n!}{a^{n+1}}.$
$$\implies S=\sum_{k=0}^{\infty} \int_{0}^{\infty}\frac{t^ke^{-(k+1)t}}{k!} dt=\sum_{k=0}^{\infty}\frac{1}{(k+1)^{k+1}}$$
A: Uniform convergence of the power series allows to interchange summation and integration:
\begin{aligned}
&\int_{0}^{1} x^{-x} \, \mathrm d x=\int_{0}^{1} e^{-x \ln x}\, \mathrm d x \Rightarrow \text { Applying Power Series of } e^{k x}=\sum_{n=0}^{\infty} \frac{(k x)^{n}}{n !} \\
&\Rightarrow \int_{0}^{1} \sum_{n=0}^{\infty} \frac{(-\ln x)^{n}}{n !} x^{n} \, \mathrm d x \Rightarrow \sum_{n=0}^{\infty} \frac{1}{n !} \int_{0}^{1} x^{n}(-\ln x)^{n} \, \mathrm d x
\end{aligned}
\begin{aligned}
&\Rightarrow\left\{u=\ln x \Leftrightarrow e^{u}=x \\ \mathrm d x=e^{u} \, \mathrm d u\right\} \Rightarrow \sum_{n=0}^{\infty} \frac{1}{n !} \int_{-\infty}^{0} e^{u(n+1)}(-u)^{n}\, \mathrm d u \\
&\Rightarrow\left\{t=-u(n+1) \Leftrightarrow u=-\frac{t}{n+1} \\ \mathrm d u=-\frac{\mathrm d t}{n+1}\right\} \Rightarrow \sum_{n=0}^{\infty} \frac{1}{n !} \int_{\infty}^{0} e^{-t}\left(\frac{t}{n+1}\right)^{n} \frac{(-1)}{n+1}\, \mathrm d t
\end{aligned}
$\Rightarrow \sum_{n=0}^{\infty} \frac{1}{n !} \int_{0}^{\infty} e^{-t} \frac{t^{n}}{(n+1)^{n+1}} \, \mathrm d t=\sum_{n=0}^{\infty} \frac{1}{n !(n+1)^{n+1}} \int_{0}^{\infty} e^{-t} t^{n} \, \mathrm d t=\sum_{n=0}^{\infty} \frac{\Gamma(n+1)}{n !(n+1)^{n+1}}=\sum_{n=1}^{\infty} \frac{1}{n^{n}}$
