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

I'm currently working on the following family of integrals: $$$$I_n = \int_0^1 \left(\frac{x - 1}{\ln(x)} \right)^n\:dx$$$$ Where $$n \in \mathbb{N}$$. I employed Feynman's Trick coupled with the Dominated Convergence Theorem and Leibniz's Integral Rule. In doing so, I introduced the following function: $$$$J_n(t) = \int_0^1 \left(\frac{x^t - 1}{\ln(x)} \right)^n\:dx$$$$ Where $$0 \leq t \leq 1 \subset \mathbb{R}$$. With some fairly easy steps, I end up with the following ODE: $$$$J_n^n(t) = (-1)^n \sum_{j = 1}^n {n \choose j} (-1)^j \frac{j^n}{jt + 1} \nonumber$$$$ Where $$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: $$$$I_n = \int_0^1 \left( \frac{x - 1}{\ln(x)} \right)^n \:dx \nonumber$$$$0 To begin with, consider the case when $$n = 1$$: $$$$I_1 = \int_0^1 \frac{x - 1}{\ln(x)}\:dx \nonumber$$$$ Here we introduce the function: $$$$J_1(t) = \int_0^1 \frac{x^t - 1}{\ln(x)}\:dx \nonumber$$$$ We observe that $$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$$: $$$$J_1'(t) = \int_0^1 \frac{\frac{d}{dt}\left[x^t - 1 \right]}{\ln(x)}\:dx = \int_0^1 \frac{\ln(x)x^t}{\ln(x)}\:dx = \int_0^1 x^t \:dx = \left[ \frac{x^{t +1}}{t + 1}\right]_0^1 = \frac{1}{t + 1} \nonumber$$$$ We now integrate with respect to $$t$$: $$$$J_1(t) = \int \frac{1}{t + 1} \:dt = \ln\left|t + 1 \right| + C \nonumber$$$$ Where $$C$$ is the constant of integration. To resolve $$C$$ we employ $$J_1(0) = 0$$: $$$$J_1(0) = 0 = \ln\left|0 + 1\right| + C \rightarrow C = 0 \nonumber$$$$ Thus, $$$$J_1(t) = \ln\left|t + 1\right| \nonumber$$$$ We now resolve $$I_1$$ using $$I_1 = J_1(1)$$: $$$$I_1 = J_1(1) = \ln\left|1 + 1\right| = \ln\left|2\right| \nonumber$$$$ The question I have is: Can this approach be used for other or all values of $$n$$?. To address this, I will proceed by applying the same method to $$n = 2$$: $$$$I_2 = \int_0^1 \frac{\left(x - 1 \right)^2}{\ln^2(x)}\:dx \nonumber$$$$ We introduce the function: $$$$J_2(t) = \int_0^1 \frac{\left( x^t - 1\right)^2}{\ln^2(x)}\:dx \nonumber$$$$ We observe that $$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$$: $$$$J_2'(t) = \int_0^1 \frac{\frac{d}{dt}\left[\left(x^t - 1\right)^2 \right]}{\ln^2(x)}\:dx = \int_0^1 \frac{2\left(x^t - 1\right)\ln(x)x^t}{\ln^2(x)}\:dx = 2 \int_0^1 \frac{x^t\left(x^t - 1\right)}{\ln(x)}\:dx \nonumber$$$$ We observe that $$J_2'(0) = 0$$. We now differentiate again with respect to $$t$$: $$$$J_2''(t) = 2\int_0^1 \frac{\ln(x)x^t\cdot \left(x^t - 1\right) + x^t \cdot \ln(x)x^t}{\ln(x)}\:dx = 2\int_0^1 2x^{2t} - x^t \:dx = 2\left[\frac{2x^{2t + 1}}{2t + 1 } - \frac{x^{t + 1}}{t + 1} \right]_0^1 = 2\left[\frac{2}{2t + 1} - \frac{1}{t + 1}\right] \nonumber$$$$ We now integrate with respect to $$t$$: $$$$J_2'(t) = 2\int \frac{2}{2t + 1} - \frac{1}{t + 1} \:dt =2\bigg[ \ln\left|2t + 1\right| - \ln\left|t + 1\right| \bigg] + C \nonumber$$$$ Where $$C$$ is the constant of integration. To resolve $$C$$, we use $$J_2'(0) = 0$$: $$$$J_2'(0) = 0 = 2\bigg[\ln\left|2\cdot 0 + 1\right| - \ln\left|0 + 1\right|\bigg] + C = 0 + C \rightarrow C = 0 \nonumber$$$$ Thus, $$$$J_2'(t) = 2\bigg[\ln\left|2t + 1\right| - \ln\left|t + 1\right|\bigg] \nonumber \nonumber$$$$ We now integrate again with respect to $$t$$: $$$$J_2(t) = 2\int \ln\left|2t + 1\right| - \ln\left|t + 1\right| \:dt = 2\bigg[\left(\frac{2t + 1}{2}\right)\bigg[ \ln\left|2t + 1\right| - 1 \bigg] - \bigg[ \left(t + 1\right)\ln\left|t + 1\right| - t \bigg] \bigg] + D \nonumber$$$$ Where $$D$$ is the constant of integration. To resolve $$D$$ we use the condition $$J_2(0) = 0$$: $$$$J_2(0) = 0 = 2\bigg[\left(\frac{2\cdot 0 + 1}{2}\right)\bigg[ \ln\left|2\cdot 0 + 1\right| - 1 \bigg] - \bigg[ \left(0 + 1\right)\ln\left|0 + 1\right| - 0 \bigg]\bigg] + D = -1+ D \rightarrow D = 1 \nonumber$$$$ Thus, $$$$J_2(t) = 2\bigg[\left(\frac{2t + 1}{2}\right)\bigg[ \ln\left|2t + 1\right| - 1 \bigg] - \bigg[ \left(t + 1\right)\ln\left|t + 1\right| - t \bigg]\bigg] + 1 \nonumber = \left(2t + 1\right)\ln\left|2t + 1\right| -2\left(t + 1\right)\ln\left|t + 1\right| \nonumber$$$$ Thus, we now may resolve $$I_2$$ using $$I_2 = J_2(1)$$: $$$$I_2 = J_2(1) = \left( 2\cdot 1 + 1\right)\ln\left|2\cdot 1 + 1\right| -2 \left(1 + 1\right)\ln\left|1 + 1\right| = 3\ln(3) -4\ln(2) \nonumber$$$$ Here I will attempt to resolve the integral in it's general form. I will employ the same approach as for $$n = 1, 2$$ and introduce the function: $$$$J_n(t) = \int_0^1 \frac{\left(x^t - 1 \right)^n}{\ln^n(x)}\:dx \nonumber$$$$ We observe that $$I_n = J_n(1)$$ and $$J_n(0) = 0$$. We begin by expanding the integrand's numerator using the Binomail Expansion: $$$$J_n(t) = \int_0^1 \frac{\sum_{j = 0}^n { n \choose j} \left(x^t\right)^j \left(-1 \right)^{n - j}}{\ln^n(x)}\:dx = (-1)^n \sum_{j = 0}^n {n \choose j} (-1)^j \int_0^1 \frac{x^{jt}}{\ln^n(x)}\:dx \nonumber$$$$ Taking the same approach as before, we now employ Leibniz's Integral Rule and differentiate $$n$$ times under the curve with respect to $$t$$: $$$$J_n^n(t) = (-1)^n \sum_{j = 0}^n {n \choose j} (-1)^j \int_0^1 \frac{\frac{d^n}{dt^n}\left[x^{jt}\right]}{\ln^n(x)}\:dx \nonumber$$$$ Here we note: $$$$\frac{d^n}{dt^n}\left[x^{jt}\right] = j^n \ln^n(x)x^{jt} \nonumber$$$$, Noting that for $$j= 0$$, the derivative is $$0$$. Thus, $$$$J_n^n(t) = (-1)^n \sum_{j = 1}^n {n \choose j} (-1)^j \int_0^1 \frac{j^n \ln^n(x)x^{jt}}{\ln^n(x)}\:dx = (-1)^n \sum_{j = 1}^n {n \choose j} (-1)^j j^n \int_0^1 x^{jt}\:dx = (-1)^n \sum_{j = 1}^n {n \choose j} (-1)^j \frac{j^n}{jt + 1} \nonumber$$$$ Where $$J_n^k(0) = 0$$ for $$k = 0,\dots, n$$.

• An integral evaluation appears in the solution to a problem proposed by Cornel I. Valean in MathProblems Journal (Volume 5, Issue 3, 2015), pages 445-447 here: mathproblems-ks.org/?wpfb_dl=80 Jul 31 '19 at 13:08
• Thanks for the reference. I have a lot of work ahead of me to understand the paper. Happy though to see this integral has served interest in minds far more sophisticated than my own!
– user679268
Jul 31 '19 at 13:35

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.

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, $$\begin{gather*} J_n(t) = \dfrac{1}{(n-1)!}\int_{0}^{\infty}\dfrac{g^{(n-1)}(y)}{y}\, dy =\\[2ex] \dfrac{1}{(n-1)!}\int_{0}^{\infty}\dfrac{g^{(n-1)}(y)-g^{(n-1)}(0)}{y}\, dy =\\[2ex] \dfrac{1}{(n-1)!}\sum_{k=0}^{n}\binom{n}{k}(-1)^{n+k-1}(kt+1)^{n-1}\int_{0}^{\infty}\dfrac{e^{-(k+1)y}-e^{-y}}{y}\, dy =\\[2ex] \dfrac{1}{(n-1)!}\sum_{k=0}^{n}\binom{n}{k}(-1)^{n+k}(kt+1)^{n-1}\ln(kt+1). \end{gather*}$$

• Nice solution! (+1) Aug 2 '19 at 9:44
• @JanG Excellent Solution - I am writing a book of integrals for my Niece and I would like to reference you, can I contact you?
– user679268
Aug 2 '19 at 11:14
• $@$Song Thank you.
– JanG
Aug 2 '19 at 12:27
• $@$Kevin Nivek You are welcome. Via email?
– JanG
Aug 2 '19 at 12:29
• @JanG - Which way you are comfortable with. My email is dp_galea@hotmail.com
– user679268
Aug 2 '19 at 22:16

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 $$J_n(t) = \left[\hat D^n [F_n](x)\right]_{x=0} = \frac 1 {(n-1)!}\sum_{k=0}^n {n \choose k} (-1)^{n-k} (kt+1)^{n-1}\ln (kt+1).$$