TITLE: Examining $\int_0^1 \left(\frac{x - 1}{\ln(x)} \right)^n\:dx$
QUESTION [13 upvotes]: I'm currently working on the following family of integrals:
\begin{equation}
 I_n = \int_0^1 \left(\frac{x - 1}{\ln(x)} \right)^n\:dx
\end{equation} 
Where $n \in \mathbb{N}$. I employed Feynman's Trick coupled with the Dominated Convergence Theorem and Leibniz's Integral Rule. In doing so, I introduced the following function:
\begin{equation}
 J_n(t) = \int_0^1 \left(\frac{x^t - 1}{\ln(x)} \right)^n\:dx
\end{equation}
Where $0 \leq t \leq 1 \subset \mathbb{R}$.  With some fairly easy steps, I end up with the following ODE:
\begin{equation}
    J_n^n(t) =  (-1)^n \sum_{j = 1}^n {n \choose j} (-1)^j \frac{j^n}{jt + 1}   \nonumber
\end{equation}
Where $J_n^k(t)$ is the $k$-th derivative of $J_n(t)$ with the conditions $J_n^k(0) = 0$ for $ 0 \leq k \leq n$. As such, to resolve $J_n(t)$ I need to integrate $J_n^n(t)$ $n$ times whilst applying the initial conditions. Although I can do it for any fixed $n$, I'm yet to be able to generalise it for any $n$. I was wondering if anyone has working with this type of ODE and if so, is there any preferable ways to approach it?
Initially I thought that using Laplace Transforms would be ideal as in applying it to $J_n^n(t)$ all terms would be removed given the initial condition. This felt apart as the Laplace Transform of $\frac{1}{t + a}$ is a nasty Special Function to work with. 
So, to repeat, is there an approach people can recommend?

For anyone who may be interested, here is my work on this integral:
In this section, I would like to address the following family of integrals:
\begin{equation}
    I_n = \int_0^1 \left( \frac{x - 1}{\ln(x)} \right)^n \:dx \nonumber 
\end{equation}0
To begin with, consider the case when $n = 1$:
\begin{equation}
    I_1 = \int_0^1 \frac{x - 1}{\ln(x)}\:dx \nonumber 
\end{equation}
Here we introduce the function:
\begin{equation}
    J_1(t) = \int_0^1 \frac{x^t - 1}{\ln(x)}\:dx \nonumber 
\end{equation}
We observe that $I_1 = J_1(1)$ and $J_1(0) = 0$. Here we employ Leibniz's  Integral Rule and differentiate under the curve with respect to $t$:
\begin{equation}
    J_1'(t) = \int_0^1 \frac{\frac{d}{dt}\left[x^t - 1 \right]}{\ln(x)}\:dx = \int_0^1 \frac{\ln(x)x^t}{\ln(x)}\:dx = \int_0^1 x^t \:dx = \left[ \frac{x^{t  +1}}{t + 1}\right]_0^1 = \frac{1}{t + 1} \nonumber 
\end{equation}
We now integrate with respect to $t$:
\begin{equation}
    J_1(t) = \int \frac{1}{t + 1} \:dt = \ln\left|t + 1 \right| + C \nonumber 
\end{equation}
Where $C$ is the constant of integration. To resolve $C$ we employ $J_1(0) = 0$:
\begin{equation}
    J_1(0) = 0 = \ln\left|0 + 1\right| + C \rightarrow C = 0 \nonumber
\end{equation}
Thus, 
\begin{equation}
    J_1(t) = \ln\left|t + 1\right| \nonumber
\end{equation}
We now resolve $I_1$ using $I_1 = J_1(1)$:
\begin{equation}
    I_1 = J_1(1) = \ln\left|1 + 1\right| = \ln\left|2\right| \nonumber 
\end{equation}
The question I have is: Can this approach be used for other or all values of $n$?. To address this, I will proceed by applying the same method to $n = 2$:
\begin{equation}
    I_2 = \int_0^1  \frac{\left(x - 1 \right)^2}{\ln^2(x)}\:dx \nonumber
\end{equation}
We introduce the function:
\begin{equation}
    J_2(t) = \int_0^1 \frac{\left( x^t - 1\right)^2}{\ln^2(x)}\:dx \nonumber
\end{equation}
We observe that $I_2 = J_2(1)$ and $J_2(0) = 0$. We proceed here by employ Leibniz's Integral Rule and differentiate under the curve with respect to $t$:
\begin{equation}
    J_2'(t) = \int_0^1 \frac{\frac{d}{dt}\left[\left(x^t - 1\right)^2 \right]}{\ln^2(x)}\:dx = \int_0^1 \frac{2\left(x^t - 1\right)\ln(x)x^t}{\ln^2(x)}\:dx = 2 \int_0^1 \frac{x^t\left(x^t - 1\right)}{\ln(x)}\:dx \nonumber 
\end{equation}
We observe that $J_2'(0) = 0$. We now differentiate again with respect to $t$:
\begin{equation}
    J_2''(t) = 2\int_0^1 \frac{\ln(x)x^t\cdot \left(x^t - 1\right) + x^t \cdot \ln(x)x^t}{\ln(x)}\:dx = 2\int_0^1 2x^{2t} - x^t \:dx = 2\left[\frac{2x^{2t + 1}}{2t + 1 } - \frac{x^{t + 1}}{t + 1} \right]_0^1 = 2\left[\frac{2}{2t + 1} - \frac{1}{t + 1}\right] \nonumber 
\end{equation}
We now integrate with respect to $t$:
\begin{equation}
    J_2'(t) = 2\int \frac{2}{2t + 1} - \frac{1}{t + 1} \:dt =2\bigg[ \ln\left|2t + 1\right| - \ln\left|t + 1\right| \bigg]  + C \nonumber 
\end{equation}
Where $C$ is the constant of integration. To resolve $C$, we use $J_2'(0) = 0$:
\begin{equation}
    J_2'(0) = 0 = 2\bigg[\ln\left|2\cdot 0 + 1\right| - \ln\left|0 + 1\right|\bigg] + C = 0 + C \rightarrow C = 0 \nonumber
\end{equation}
Thus, 
\begin{equation}
    J_2'(t) = 2\bigg[\ln\left|2t + 1\right| - \ln\left|t + 1\right|\bigg] \nonumber \nonumber 
\end{equation}
We now integrate again with respect to $t$:
\begin{equation}
    J_2(t) = 2\int  \ln\left|2t + 1\right| - \ln\left|t + 1\right| \:dt = 2\bigg[\left(\frac{2t + 1}{2}\right)\bigg[ \ln\left|2t + 1\right| - 1 \bigg] - \bigg[ \left(t + 1\right)\ln\left|t + 1\right| - t \bigg] \bigg] + D \nonumber 
\end{equation}
 Where $D$ is the constant of integration. To resolve $D$ we use the condition $J_2(0) = 0$:
 \begin{equation}
     J_2(0) = 0 = 2\bigg[\left(\frac{2\cdot 0 + 1}{2}\right)\bigg[ \ln\left|2\cdot 0 + 1\right| - 1 \bigg] - \bigg[ \left(0 + 1\right)\ln\left|0 + 1\right| - 0 \bigg]\bigg] + D   = -1+ D \rightarrow D = 1 \nonumber 
 \end{equation}
 Thus, 
 \begin{equation}
     J_2(t) = 2\bigg[\left(\frac{2t + 1}{2}\right)\bigg[ \ln\left|2t + 1\right| - 1 \bigg] - \bigg[ \left(t + 1\right)\ln\left|t + 1\right| - t \bigg]\bigg] + 1 \nonumber = \left(2t + 1\right)\ln\left|2t + 1\right| -2\left(t  + 1\right)\ln\left|t + 1\right| \nonumber 
 \end{equation}
 Thus, we now may resolve $I_2$ using $I_2 = J_2(1)$:
 \begin{equation}
     I_2 = J_2(1) = \left( 2\cdot 1 + 1\right)\ln\left|2\cdot 1 + 1\right| -2 \left(1 + 1\right)\ln\left|1 + 1\right| = 3\ln(3) -4\ln(2) \nonumber 
\end{equation}
Here I will attempt to resolve the integral in it's general form. I will employ the same approach as for $n = 1, 2$ and introduce the function:
\begin{equation}
    J_n(t) = \int_0^1 \frac{\left(x^t - 1 \right)^n}{\ln^n(x)}\:dx \nonumber 
\end{equation}
We observe that $I_n = J_n(1)$ and $J_n(0) = 0$. We begin by expanding the integrand's numerator using the Binomail Expansion:
\begin{equation}
    J_n(t) = \int_0^1 \frac{\sum_{j = 0}^n { n \choose j} \left(x^t\right)^j \left(-1 \right)^{n - j}}{\ln^n(x)}\:dx = (-1)^n \sum_{j = 0}^n {n \choose j} (-1)^j \int_0^1 \frac{x^{jt}}{\ln^n(x)}\:dx \nonumber
\end{equation}
Taking the same approach as before, we now employ Leibniz's Integral Rule and differentiate $n$ times under the curve with respect to $t$:
\begin{equation}
    J_n^n(t) = (-1)^n \sum_{j = 0}^n {n \choose j} (-1)^j \int_0^1 \frac{\frac{d^n}{dt^n}\left[x^{jt}\right]}{\ln^n(x)}\:dx \nonumber
\end{equation}
Here we note:
\begin{equation}
    \frac{d^n}{dt^n}\left[x^{jt}\right] = j^n \ln^n(x)x^{jt} \nonumber 
\end{equation}, 
Noting that for $j= 0$,  the derivative is $0$. Thus, 
\begin{equation}
    J_n^n(t) = (-1)^n \sum_{j = 1}^n {n \choose j} (-1)^j \int_0^1 \frac{j^n \ln^n(x)x^{jt}}{\ln^n(x)}\:dx = (-1)^n \sum_{j = 1}^n {n \choose j} (-1)^j j^n \int_0^1 x^{jt}\:dx  =  (-1)^n \sum_{j = 1}^n {n \choose j} (-1)^j \frac{j^n}{jt + 1}   \nonumber
\end{equation}
Where $J_n^k(0) = 0$ for $k = 0,\dots, n$.

REPLY [4 votes]: With
\begin{equation*}
 I_n = \int_{0}^{1}\left(\dfrac{x-1}{\ln(x)}\right)^n\, dx
\end{equation*}
and the substitution $x=e^{-y}$  we get that
\begin{equation*}
 I_n= \int_{0}^{\infty}\dfrac{f(y)}{y^n}\, dy
\end{equation*}
where 
\begin{equation*}
 f(y) = \left(1-e^{-y}\right)^ne^{-y} =\sum_{k=0}^{n}\binom{n}{k}(-1)^{k}e^{-(k+1)y}.
\end{equation*}
Then $f(0)=0$ and $y=0$ is a zero of order $n$. Consequently $f^{(n-1)}(0)=0$. But
\begin{equation*}
 f^{(n-1)}(0)=\sum_{k=0}^{n}\binom{n}{k}(-1)^{n+k-1}(k+1)^{n-1}=0.\tag{1}
\end{equation*}
Now we are prepared to evaluate $I_n$. After integration by parts $n-1$ times we have
\begin{gather*}
 I_n = \dfrac{1}{(n-1)!}\int_{0}^{\infty}\dfrac{f^{(n-1)}(y)}{y}\, dy =\\[2ex] \dfrac{1}{(n-1)!}\int_{0}^{\infty}\dfrac{1}{y}\sum_{k=0}^{n}\binom{n}{k}(-1)^{n+k-1}(k+1)^{n-1}e^{-(k+1)y}\, dy.
\end{gather*}
However, if we use $(1)$ we get
\begin{gather*}
 I_n =  \dfrac{1}{(n-1)!}\sum_{k=0}^{n}\binom{n}{k}(-1)^{n+k-1}(k+1)^{n-1}\int_{0}^{\infty}\dfrac{e^{-(k+1)y}-e^{-y}}{y}\, dy =\\[2ex]
 \dfrac{1}{(n-1)!}\sum_{k=0}^{n}\binom{n}{k}(-1)^{n+k}(k+1)^{n-1}\ln(k+1)
\end{gather*}
where we in the last step have used Frullani's integral. 
See https://en.wikipedia.org/wiki/Frullani_integral 
The integral
\begin{equation*}
 J_n(t) = \int_{0}^{1}\left(\dfrac{x^t-1}{\ln(x)}\right)^{n}\, dx =
\end{equation*}
can be treated similarly to $I_n$.
\begin{equation*}
J_n(t) =\int_{0}^{\infty}\dfrac{g(y)}{y^n}\, dy
\end{equation*}
where 
\begin{equation*}
 g(y) = \left(1-e^{-ty}\right)^ne^{-y} =\sum_{k=0}^{n}\binom{n}{k}(-1)^{k}e^{-(kt+1)y}.
\end{equation*}
Furthermore,
\begin{gather*}
 J_n(t) =  \dfrac{1}{(n-1)!}\int_{0}^{\infty}\dfrac{g^{(n-1)}(y)}{y}\, dy =\\[2ex]
\dfrac{1}{(n-1)!}\int_{0}^{\infty}\dfrac{g^{(n-1)}(y)-g^{(n-1)}(0)}{y}\, dy =\\[2ex]
\dfrac{1}{(n-1)!}\sum_{k=0}^{n}\binom{n}{k}(-1)^{n+k-1}(kt+1)^{n-1}\int_{0}^{\infty}\dfrac{e^{-(k+1)y}-e^{-y}}{y}\, dy =\\[2ex]
\dfrac{1}{(n-1)!}\sum_{k=0}^{n}\binom{n}{k}(-1)^{n+k}(kt+1)^{n-1}\ln(kt+1).
\end{gather*}