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 $k = 0,\dots, n$.
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