Examining $\int_0^1 \left(\frac{x - 1}{\ln(x)} \right)^n\:dx$

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 $k = 0,\dots, n$.

The general formula for $I_n$ is $$ I_n = \frac 1 {(n-1)!}\sum_{k=1}^n {n \choose k} (-1)^{n-k} (k+1)^{n-1} \ln (k+1) $$ and for $J_n(t)$, we have $$ J_n(t) = \frac 1 {(n-1)!}\sum_{k=1}^n {n \choose k} (-1)^{n-k} (kt+1)^{n-1} \ln (kt+1). $$ I've checked that this formula returns correct values \begin{align*} I_1 =& \ln 2\\ I_2 =& - 4\ln 2 + 3\ln 3\\ I_3 =& 22\ln 2 -\frac{27}{2}\ln 3\\ I_4 =& -\frac{272}{3} \ln 2 + 27\ln 3 +\frac {125}6 \ln 5. \end{align*} Evaluation of $I_n$: To see this, we first make change of variable $ y = -\ln x$ to find that \begin{align*} I_n = & \int_0^\infty \left(\frac{1-e^{-y}}{y}\right)^n e^{-y} \mathrm dy \\ =& \int_0^\infty \left(\int_0^1 e^{-vy} \mathrm dv\right)^n e^{-y} \mathrm dy \\ =& \int_0^\infty \int_0^1 \cdots \int_0^1 e^{-y(1+v_1+v_2 + \cdots +v_n)} \mathrm dv_1\cdots \mathrm d v_n \mathrm dy \\ =& \int_0^1 \cdots \int_0^1 \frac 1 { 1+ v_1 + \cdots +v_n} \mathrm dv_1 \cdots \mathrm dv_n. \end{align*} To calculate $I_n$ iteratively, let us define \begin{align*} F_n(x) := \int_0^x \int_0^{x_{n-1}} \cdots \int_0^{x_1} \frac 1 {1+t} \mathrm dt \mathrm dx_1\cdots \mathrm dx_{n-1} . \end{align*} We can find that for $n\ge 1$, $$ F_n(x) = \frac {(x+1)^{n-1}}{(n-1)!}\ln (x+1) - \frac{H_{n-1}}{(n-1)!}(x+1)^{n-1} $$ where $ H_n = 1+ \frac 1 2 + \cdots + \frac 1n$ is the $n$-th harmonic number. (I skipped the derivation, but given the form we can at least check that $F_n' = F_{n-1}$ and $F_1(x) = \ln (x+1)$. So mathematical induction can be applied.)

For convenience let us introduce the noation $F$ for the forward opeartor $ \displaystyle F[f](x) = f(x+1) $ and the forward difference operator $ \displaystyle D[f](x) = F[f](x) - I[f](x) = f(x+1)- f(x) $ where $f$ is an arbitrary function. Note that these are linear operators, and we write $D^n = (F-I)^n $ as the iteration of $D$. Now, since $\displaystyle \int D[f] = D\left[\int f\right] + C$, we can integrate $I_n$ iteratively; \begin{align*} I_n = &\int_0^1 \cdots \int_0^1 \frac 1 { 1+ v_1 + \cdots +v_n} \mathrm dv_1 \cdots \mathrm dv_n\\ =& \int_0^1 \cdots \int_0^1 \left[F_1(v_1+v_2+\cdots +v_n)\right]^{v_1=1}_{v_1=0}\mathrm dv_2 \cdots \mathrm dv_n\\ =& \int_0^1 \cdots \int_0^1 D[F_1](v_2+\cdots +v_n)\mathrm dv_2 \cdots \mathrm dv_n \\ =& \int_0^1 \cdots \int_0^1 \left[D[F_2](v_2+\cdots +v_n)\right]^{v_2=1}_{v_2=0}\mathrm dv_3 \cdots \mathrm dv_n \\ =& \int_0^1 \cdots \int_0^1 D^2[F_2](v_3+\cdots +v_n) \mathrm dv_3 \cdots \mathrm dv_n \\ =& \cdots \int_0^1 D ^{n-1} [F_{n-1}](v_n)\mathrm dv_n\\ =& \left[ D^{n-1} [F_n](v_n)\right]^{v_n=1}_{v_n=0}\\ =& D^n[F_n](0). \end{align*} We notice that $\displaystyle D^k[x^{k-j}] \equiv 0$ for $j\ge 1$, i.e. polynomials of degree less than $k$ becomes $0$ when $k$-times differenced (because its degree decreases by $1$ each time it is differenced.) So we have \begin{align*} D^n [F_n](x) =& D^n\left [\frac {(x+1)^{n-1}}{(n-1)!}\ln (x+1) - \frac{H_{n-1}}{(n-1)!}(x+1)^{n-1}\right] \\ =& D^n\left [\frac {(x+1)^{n-1}}{(n-1)!}\ln (x+1)\right] -0 \\ =& (F-I)^n \left [\frac {(x+1)^{n-1}}{(n-1)!}\ln (x+1)\right] \\\\ =& \sum_{k=0}^n {n\choose k} F^k (-I)^{n-k} \left [\frac {(x+1)^{n-1}}{(n-1)!}\ln (x+1)\right]\\ =& \sum_{k=0}^n {n \choose k} (-1)^{n-k} \frac {(x+k+1)^{n-1}}{(n-1)!}\ln (x+k+1). \end{align*} Therefore, it follows that $$ I_n = D^n[F_n](0) = \frac 1 {(n-1)!}\sum_{k=1}^n {n\choose k} (-1)^{n-k} (k+1)^{n-1} \ln (k+1). $$ Addendum, Evaluation of $J_n(t)$: The same change of variable $y = -\ln x$ gives us that $$ J_n(t) = \int_0^t \cdots \int_0^t \frac 1 { 1+v_1 + \cdots +v_n} \mathrm dv_1 \cdots \mathrm dv_n. $$ Nothing really changes except that we now define $t$-step forward and forward difference as \begin{align*} \hat F[f](x) =& f(x+t)\\ \hat D[f](x) = & \hat F[f](x) - I[f](x) = f(x+t) - f(x). \end{align*} Then, \begin{align*} J_n(t) = &\int_0^t \cdots \int_0^t \frac 1 { 1+ v_1 + \cdots +v_n} \mathrm dv_1 \cdots \mathrm dv_n\\ =& \int_0^t \cdots \int_0^t \left[F_1(v_1+v_2+\cdots +v_n)\right]^{t}_{v_1=0}\mathrm dv_2 \cdots \mathrm dv_n\\ =& \int_0^t \cdots \int_0^t \hat D[F_1](v_2+\cdots +v_n)\mathrm dv_2 \cdots \mathrm dv_n \\ =& \int_0^t \cdots \int_0^t \left[\hat D[F_2](v_2+\cdots +v_n)\right]^{t}_{v_2=0}\mathrm dv_3 \cdots \mathrm dv_n \\ =& \int_0^t \cdots \int_0^t \hat D^2[F_2](v_3+\cdots +v_n) \mathrm dv_3 \cdots \mathrm dv_n \\ =& \cdots = \int_0^t \hat D ^{n-1} [F_{n-1}](v_n)\mathrm dv_n\\ =& \left[\hat D^{n-1} [F_n](v_n)\right]^t_{v_n=0}\\ =& \hat D^n[F_n](0). \end{align*} Since \begin{align*} \hat D^n [F_n](x) =&\sum_{k=0}^n {n \choose k} \hat F^k (-I)^{n-k} \frac {(x+1)^{n-1}}{(n-1)!}\ln (x+1)\\ =&\sum_{k=0}^n {n \choose k} (-1)^{n-k} \frac {(x+kt+1)^{n-1}}{(n-1)!}\ln (x+kt+1) \end{align*} it follows $$ J_n(t) = \left[\hat D^n [F_n](x)\right]_{x=0} = \frac 1 {(n-1)!}\sum_{k=0}^n {n \choose k} (-1)^{n-k} (kt+1)^{n-1}\ln (kt+1). $$

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*}