# How can you prove this equality?

I am trying to figure out the following equality, but cannot seem to get anywhere. I tried integrating by parts, but that blew up when I set u = (log x)^n and tried to take log (0). I also tried differentiating the right side but got stuck when I did not know the derivative of $n!$

How can we prove that the following is true:

$$\int_0^1 x^a(\log x)^n \,dx= \frac{(-1)^n(n!)}{(a+1)^{n+1}}$$

Any suggestions would be greatly appreciated.

• These kind of things usually succumb to integration by parts and then induction. – marty cohen Jan 29 '14 at 2:19
• (1) $\log x\to -\infty$ as $x\to 0$, anyway, so this was an improper integral in the first place, so you should expect that sort of behavior; (2) $n!$ is a constant (if we're differentiating with respect to $x$), so its derivative is zero; (3) Since we're dealing with a definite integral, the whole right side is a constant, so we can't be taking a derivative anyway. – tabstop Jan 29 '14 at 2:19

Applying integration by parts:

$$\int_0^1 x^a(log x)^ndx = \lim_{b\to 0^+}\left[(\log(x))^{n}\frac{1}{a+1}x^{a+1}\Big|_{b}^1\right]-\frac{n}{a+1}\int_0^1 x^{a}(\log x)^{n-1}dx$$

Now if $a>-1$, you can show with L'Hopital's rule that

$$\lim_{b\to 0^+}\left[(\log(b))^{n}b^{a+1}\right] = 0$$

Hence

$$\int_0^1 x^a(log x)^ndx = -\frac{n}{a+1}\int_0^1 x^{a}(\log x)^{n-1}dx$$

See the inductive pattern now? Why you will have $n$ factors of $(-1)$ when you're finished?

Note that $x^a \log^n(x) = \dfrac{\partial^n}{\partial a^n} x^a$, so you just need to start with $\int_0^1 x^a\ dx$ and take some derivatives. Of course you do need $a > -1$ for this to converge.

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{1}x^{a}\ln^{n}\pars{x}\,\dd x = {\pars{-1}^{n}\,n! \over \pars{a + 1}^{n+1}}:\ {\large ?}}$

With $\ds{a > -1}$: \begin{align} \color{#00f}{\large\int_{0}^{1}x^{a}\ln^{n}\pars{x}\,\dd x}&= \lim_{\mu \to 0}\partiald[n]{}{\mu}\int_{0}^{1}x^{a + \mu}\,\dd x = \lim_{\mu \to 0}\partiald[n]{}{\mu}\pars{1 \over a + \mu + 1} = \partiald[n]{}{a}\pars{1 \over a + 1} \\[3mm]&= \partiald[n - 1]{}{a}\pars{-1 \over \pars{a + 1}^{2}} = \partiald[n - 2]{}{a}\pars{2 \over \pars{a + 1}^{3}} = \partiald[n - 3]{}{a}\pars{-3\times 2 \over \pars{a + 1}^{4}} \\[3mm]&=\cdots= \partiald[n - k]{}{a}\pars{\pars{-1}^{k}k! \over \pars{a + 1}^{k + 1}} =\cdots=\color{#00f}{\large{\pars{-1}^{n}\,n! \over \pars{a + 1}^{n + 1}}} \end{align}