How do you show that $\int_{0}^1\frac{dx}{x^x}=\sum_{k=1}^\infty\frac{1}{k^k}$? My task is this:
i) Find the sum to$$1-x\ln x +\frac{1}{2}(x\ln x)^2-\ldots+\frac{(-1)^k}{k!}(x\ln x)^k+\ldots$$
(ii) The great norwegian mathematician Atle Selberg showed that $$\int_{0}^1\frac{dx}{x^x}=\sum_{k=1}^\infty\frac{1}{k^k}$$ when he was 15. Can you?
My works so far:
By inspection we can relate (i) to a well known series, namely $$\begin{align}e^x=\sum_{n=0}^\infty\frac{x^n}{n!}= 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\ldots+\frac{x^n}{n!}+\dots\\\implies e^{-x\ln(x)}=e^{\ln x^{-x}}=x^{-x}= \frac{1}{x^x}=\\\sum_{n=0}^\infty\frac{\big((-x\ln x\big)^n}{n!}=\sum_{n=0}^\infty\frac{(-1)^n(x\ln x)^n}{n!}=\\1-x\ln x +\frac{1}{2}(x\ln x)^2-\ldots+\frac{(-1)^k}{k!}(x\ln x)^k+\ldots\end{align}$$
Which is what we wanted to show. Now comes the hard part and sadly I can't contribute much. I did tried this for fun $$\begin{align}\left(e^{-x\ln x}\right)'=e^{-x\ln x}(-x\ln x)'=e^{-x\ln x}(-\ln x - 1)=-\frac{\ln x +1}{x^x}\\\frac{d}{dx}\left[\int\frac{dx}{x^x}\right]=\frac{1}{x^x}=\frac{d}{dk}\left[\sum_{n=1}^\infty\frac{1}{k^k}\right]=-\sum_{n=1}^\infty \frac{\ln k +1}{k^k}\end{align}$$ 
Sadly I can't see how this is useful in order to evaluate the integral. Any help would be appriciated. Thanks in advance!
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,}
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 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
\color{#f00}{\int_{0}^{1}{\dd x \over x^{x}}} & =
\int_{0}^{1}\expo{-x\ln\pars{x}}\,\dd x =
\sum_{n = 0}^{\infty}{\pars{1}^{n} \over n!}\int_{0}^{1}x^{n}\ln^{n}\pars{x}\,\dd x\tag{1}
\end{align}
Also,
\begin{align}
\color{#00f}{\int_{0}^{1}x^{a}\,\dd x} & = {1 \over a + 1}\quad\imp\quad
\left\lbrace\begin{array}{lcr}
\ds{\int_{0}^{1}x^{a}\ln\pars{x}\,\dd x} & \ds{=} &
\ds{-\,{1 \over \pars{a + 1}^{2}}}
\\
\ds{\int_{0}^{1}x^{a}\ln^{2}\pars{x}\,\dd x} & \ds{=} &
\ds{{2 \over \pars{a + 1}^{3}}}
\\
\ds{\int_{0}^{1}x^{a}\ln^{3}\pars{x}\,\dd x} & \ds{=} &
\ds{-\,{3.2 \over \pars{a + 1}^{4}}}
\\
\vdots & = & \vdots
\end{array}\right.
\\[3mm] \mbox{such that}\ \int_{0}^{1}x^{a}\ln^{n}\pars{x}\,\dd x & =
\pars{-1}^{n}\,{n\pars{n - 1}\ldots 2.1 \over \pars{a + 1}^{n + 1}} =
\pars{-1}^{n}\,{n! \over \pars{a + 1}^{n + 1}}
\\[3mm]
\mbox{and with}\ a \to n; &\quad 
\int_{0}^{1}x^{n}\ln^{n}\pars{x}\,\dd x =
\pars{-1}^{n}\,{n! \over \pars{n + 1}^{n + 1}}
\end{align}
Replace this result in $\pars{1}$:
$$
\color{#f00}{\int_{0}^{1}{\dd x \over x^{x}}} =
\color{#f00}{\sum_{n = 1}^{\infty}{1 \over n^{n}}}
$$
