Evaluate the definite integral $\int_0^1 {\frac{{\left( {{\rm 1 - x}} \right)^n }} {\ln x}} dx$ god moorning
I'd like to prove that
$$
\int\limits_0^1 {\frac{{\left( {{\rm 1 - x}} \right)^n }}{{\ln x}}dx = \left( {n - 1} \right)} !\int\limits_0^1 {\frac{{dx}}{{\prod\limits_{m = 1}^n {\left( {x + m} \right)} }}}  = \left( {n - 1} \right)!f\sum\limits_{m = 1}^n {\left( { - 1} \right)^m \frac{{\ln \left( {\frac{{m + 1}}{m}} \right)}}{{\left( {m - 1} \right)!\left( {n - m} \right)!}}} 
$$
 A: Notice $$\int_0^1 x^t dt = \int_0^1 e^{(\log x) t} dt = \left[\frac{e^{(\log x)t}}{\log x}\right]_0^1 = \left[\frac{x^t}{\log x}\right]_0^1 = -\frac{1-x}{\log x}$$
We have 
$$\begin{align}
\int_0^1 \frac{(1-x)^n}{\log x}dx
&= -\int_0^1 \int_0^1 (1-x)^{n-1}x^t dt dx
= -\int_0^1 \int_0^1 (1-x)^{n-1}x^t dx dt\\
&= -\int_0^1 \frac{\Gamma(n)\Gamma(t+1)}{\Gamma(t+n+1)} dt
\end{align}
$$
Notice $\Gamma(t+n+1) = \Gamma(t+1) \prod\limits_{m=1}^n (t+m)$, we get
$$\int_0^1 \frac{(1-x)^n}{\log x}dx = -\Gamma(n)\int_0^1 \frac{1}{\prod\limits_{m=1}^n(t+m)} dt \tag{*1}$$
Up to a sign, this is the first half of the equality in the question.
Since all the roots in the denominator in the RHS of $(*1)$ is simple, we can read off its partial fraction decomposition as
$$\frac{1}{\prod\limits_{m=1}^n (t+m)} 
= \sum_{m=1}^n \frac{1}{\prod\limits_{k=1,k\ne m}^n (-m + k)}\frac{1}{t+m}
= \sum_{m=1}^n \frac{(-1)^{m-1}}{(m-1)!(n-m)!}\frac{1}{t+m}$$
and the integral becomes
$$
\begin{align}
\text{RHS}(*1) 
&= -\Gamma(n)\sum_{m=1}^n \frac{(-1)^{m-1}}{(m-1)!(n-m)!} \int_0^1 \frac{1}{t+m} dt\\
&= (n-1)!\sum_{m=1}^n \frac{(-1)^m}{(m-1)!(n-m)!}\log\left(\frac{m+1}{m}\right)
\end{align}$$
Up to a sign, this is the second half of the equality in question. In short, the statement in question is basically correct (up to a missing minus sign in the middle integral).
A: Put (for $0\leq x<1$): $\displaystyle \;F(x)=\int_0^x \frac{dt}{\log t}\;$ and $\displaystyle \;G(x)=\int_0^x \left(\frac{1}{\log t}-\frac{1}{t-1}\right)\,dt$.
We see that $G$ has a finite limit $G(1)$ as $x\to 1$, and that $F(x)=G(x)+\log(1-x).$
Now for $0<h<1,$
$$I_n(h) \;=\; \int_0^h \frac{(1-x)^n}{\log x}\,dx \;=\; \sum_{k=0}^n (-1)^k \binom{n}{k}\int_0^h \frac{x^k}{\log x}\,dx.$$
We have
$$ \int_0^h \frac{x^k}{\log x} dx = \int_0^h \frac{(k+1)x^k}{\log x^{k+1}} dx = \int_0^{h^{k+1}}\frac{du}{\log u} = F(h^{k+1})=G(h^{k+1})+\log(1-h^{k+1}).$$
As $\displaystyle \log(1-h^{k+1})=\log(1-h)+\log(1+h+\cdots +h^k)$ we have
$$I_n(h)=A_n(h)+B_h(h)+C_n(h),$$
with
$$A_n(h)=\sum_{k=0}^n (-1)^k\binom{n}{k}G(h^{k+1}).$$
As $G(x)\to G(1)$ if $x\to 1$, we get that $\displaystyle A_n(h)\to G(1)\left(\sum_{k=0}^n (-1)^k \binom{n}{k}\right)=0$ if $h\to 1$.
$$B_n(h)= \sum_{k=0}^n (-1)^k\binom{n}{k}\log(1-h)=0$$
and
$$C_n(h)=\sum_{k=0}^n (-1)^k\binom{n}{k}\log(1+h+\cdots+h^k) \to \sum_{k=0}^n (-1)^k\binom{n}{k}\log(k+1)$$ if $h\to 1$.
Hence $\displaystyle I_n(h)\to \sum_{k=0}^n (-1)^k\binom{n}{k}\log(k+1)=\sum_{k=1}^n (-1)^k\binom{n}{k}\log(k+1)=I_n$ if $h\to 1$.
The other formula seems to follow from this, with some computations ($\displaystyle \int_0^1\frac{dx}{x+k}=\log(k+1)$ for $k\geq 1$).
