How to compute $\int_0^1 [\ln(1/x)]^5 dx$ Why is $$\int_0^1 [\ln(\frac{1}{x})]^5  dx=120$$? More generally,  is $$\int_0^1 [\ln(1/x)]^n  dx$$ equal to n!? If so why?
 A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{}$
\begin{align}
\int_{0}^{1}x^{\mu}\,\dd x&={1 \over \mu + 1}\quad\imp\quad
-\int_{0}^{1}x^{\mu}\ln^{5}\pars{x}\,\dd x=-\,\totald[5]{}{\mu}\bracks{1 \over \mu + 1}
={5! \over \pars{\mu + 1}^{6}}
\end{align}

Set $\ds{\mu = 0}$:
  $$\boxed{\vphantom{\Huge {A \over B}}\color{#00f}{\displaystyle\large%
\quad\int_{0}^{1}\ln^{5}\pars{1 \over x} = 120}\quad}
$$

A: So we want to find $\int_0^1(-1)^n(\ln x)^n\,dx$. Integrate by parts, letting $u=(-1)^n(\ln x)^n$ and $dv=dx$. Then we have $du=(-1)^n\frac{n}{x}(\ln x)^{n-1}\,dx$ and we can take $v=x$. Thus
$$\int_0^1 (\ln x)^n \,dx =\left.x(-1)^n(\ln x)^n\right|_0^1 +n\int_0^1(-1)^{n-1}(\ln x)^{n-1}\,dx .$$
If $I_n$ is the integral of $(-1)^n(\ln x)^n$ from $0$ to $1$, we obtain the recurrence $I_n=nI_{n-1}$.
The above recurrence, and induction, yield the desired result. 
Remark: In principle, we are dealing with an improper integral, and instead of calculating the integral from $0$ to $1$, we should calculate the integral from $\epsilon$ to $1$, and then take the limit as $\epsilon$ approaches $0$ through positive values.
A: We can solve this integral by using the log rule $\ln(a/b)=\ln a-\ln b$.  I first do the case you want, and than the general case.  We obtain $$
I\equiv \int_0^1 \ln^5\big(\frac{1}{x}\big)dx=\int_0^1 \big(\ln 1-
\ln x\big)^5 dx=\int_0^1 dx(-1)^5\ln^5x=-\int_0^1dx \ln^5 x=120.
$$
In general
$$
I\equiv \int_0^1 \ln^n\big(\frac{1}{x}\big)dx=\int_0^1 \big(\ln 1-
\ln x\big)^n dx=\int_0^1 dx(-1)^n\ln^nx =\Gamma(n+1)=n!, \ \Re(n)>-1.
$$
A: Another solution: make the change of variables $x = \exp -y$:
$$
u_n = \int_0^1 \left(\log \frac 1x\right)^n dx
= \int_0^\infty y^n e^{-y} du = n!
$$because we all know the properties of the 
$\Gamma$ function.

In case you do not, this one is just an integration by parts and an induction:
$$
u_n = \int_0^\infty y^n e^{-y} du =
[-e^{-y} y^n]_0^\infty + \int_0^\infty ny^{n-1} e^{-y} du = nu_{n-1};\\
u_0 = \int_0^\infty e^{-y} du = [-e^{-y} ]_0^\infty = 1.
$$
A: Prove it inductively!
$$\int_0^1 \ln x dx= (x\ln x -x)\bigg|_0^1=-1$$
Claim: 
$$\int_0^1 \ln^nxdx=(-1)^nn!$$
Assume the claim is true for $n=k$. Then for $n=k+1$ we have
$$\int_0^1\ln^{k+1}xdx=x\ln^{k+1}x\bigg|_0^1- \int_0^1 x (k+1) \ln^kx \frac{1}{x}dx$$
$$= -(k+1)\int_0^1\ln^kxdx=-(k+1)\cdot(-1)^k k!=(-1)^{k+1}(k+1)!$$
so the induction is finished.
Then write $(\ln(1/x))^n= (-\ln x)^n=(-1)^n (\ln x)^n$ and you have what you want.
