How can I find who discovered this integral? I need to find the first paper/author to document this integral $$\int\log^nx\;\mathrm dx=(-1)^n\;\Gamma(n+1,-\log x)\quad n\in\Bbb N_0$$
To prevent this in the future, is there a service in which I can browse integrals and view their authors? I could not find the above integral in some of the popular integral tables.
 A: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\down}{\downarrow}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
\color{#00f}{\large\int\ln^{n}\pars{x}\,\dd x}&=\lim_{\mu \to 0}\totald[n]{}{\mu}\int x^{\mu}\,\dd x
=\lim_{\mu \to 0}\totald[n]{}{\mu}\bracks{x^{\mu + 1}\pars{\mu + 1}^{-1}}
\\[3mm]&=\lim_{\mu \to 0}\sum_{k = 0}^{n}{n \choose k}\totald[n - k]{\,x^{\mu + 1}}{\mu}\,
\totald[k]{\pars{\mu + 1}^{-1}}{\mu}
\\[3mm]&=\lim_{\mu \to 0}\sum_{k = 0}^{n}{n \choose k}x^{\mu + 1}\ln^{n - k}\pars{x}\,
{\pars{-1}^{k}k! \over \pars{\mu + 1}^{k + 1}}
=\sum_{k = 0}^{n}{n \choose k}x\ln^{n - k}\pars{x}\pars{-1}^{k}k!
\\[3mm]&=
\color{#00f}{\large n!\,x\sum_{k = 0}^{n}{\pars{-1}^{k} \over \pars{n - k}!}\,
\ln^{n - k}\pars{x}}\ +\ \mbox{a constant}
\end{align}
