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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!

Thomas
  • 781

1 Answers1

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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

\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 $\pars{1}$:

$$ \color{#f00}{\int_{0}^{1}{\dd x \over x^{x}}} = \color{#f00}{\sum_{n = 1}^{\infty}{1 \over n^{n}}} $$

Felix Marin
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