1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ These kind of things usually succumb to integration by parts and then induction. $\endgroup$ – marty cohen Jan 29 '14 at 2:19
  • $\begingroup$ (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. $\endgroup$ – tabstop Jan 29 '14 at 2:19
1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.