Finding the nth integral of $\ln(1+z/k)$

Question

I can find the nth integral of $\ln(z)$ as follows: \begin{aligned} \left(\frac d{dz}\right)^{-n}\ln(z)&=\frac1{\Gamma(n)}\int\limits_0^z(z-t)^{n-1}\ln(t)dt\\ &=\frac1{n!}\left[\int\limits_0^z\frac1t(z-t)^ndt-z^n\ln(0)\right]\\ &=\frac1{n!}\left[\int\limits_0^1\frac{z^nt^n-z^n+z^n}{1-t}dt-z^n\ln(0)\right]\\ &=\frac{\ln(z)-H_n}{n!}z^n, \end{aligned} but I can't get very far with $\ln(1+z/k)$. I was able to come up with this but it is only a conjecture: \begin{aligned} \frac1{\Gamma(n)}\int\limits_0^z(z-t)^{n-1}\ln\left(1+\frac tk\right)dt&=\frac{\ln\left(1+\frac{z}{k}\right)-H_n}{n!}(k+z)^n+\sum ^{n}_{i=0}\frac{H_{i}}{i!(n-i)!}z^{n-i}k^i\\ &=\frac1{n!}\ln\left( 1+\frac{z}{k}\right)( k+z)^{n} -\sum ^{n}_{i=0}\frac{H_{n} -H_{i}}{i!( n-i) !} z^{n-i} k^{i}. \end{aligned} I'm only using $k,n\in\mathbb{N}$ and $z\in\mathbb{R}$ for now. Any help proving this would be appreciated.

Answer 1

This is not a proof

Consider $$f_1=\int_0^z \log \left(1+\frac{z}{k}\right)\,dz\qquad \text{and} \qquad f_{n}=\int_0^z f_{n-1}\,dz$$ Define $$a_n=n\, a_{n-1} \qquad \text{with} \qquad a_1=1$$ and $$g_n=\frac{a_n} z \left(f_n-\frac{(k+z)^n }{n!}\log \left(1+\frac{z}{k}\right)\right)+k^{n-1}$$

The terms $g_n$ appears to be $$g_n=-\frac { z\,P_{n-2}} {\text{lcm}(1,2,\cdots,n)} $$ where the polynomials $P$ are homogeneous is $k$,$z$, all coefficients being positive. None of these polynomials can be factored.

Tested up to extremely large values of $n$, you conjecture is verified.

In any manner, if $$I_n=\frac1{\Gamma(n)}\int\limits_0^z(z-t)^{n-1}\log\left(1+\frac tk\right)dt$$ $$I_n=\frac{z^{n+1} }{n (n+1) (z+k) \Gamma(n)}\,\, _2F_1\left(1,n+1;n+2;\frac{z}{z+k}\right)$$

Answer 2

This is not a derivation of the integral, but rather a proof of your conjecture.

Your expression is $$f_n(z)=\frac1{n!}\ln\left( 1+\frac{z}{k}\right)( k+z)^{n} -\sum ^{n}_{i=1}\frac{H_{n} -H_{n-i}}{i!( n-i) !} k^{n-i} z^{i}$$

I reversed the order of summation from yours and then got rid of the first term since $H_n - H_n = 0$. Then there are two things that need to be shown for each $n \ge 1$: $f_{n-1}(z) = \frac{d}{dz}f_n(z)$ and $f_n(z) = 0$. The second is true since each term is $0$ when $z = 0$. For the first, the derivative of $f_n(z)$ is $$\frac{1}{n!}\left( \ln\left( 1+\frac{z}{k}\right)\frac{d}{dz}(k+z)^n + (k+z)^n\frac{d}{dz}\ln\left( 1+\frac{z}{k}\right) \right) - \sum_{i=1}^n \frac{H_{n} -H_{n-i}}{i!( n-i) !} k^{n-i}\frac{d}{dz}z^{i}$$

This simplifies to $$\frac{1}{(n-1)!}\ln\left( 1+\frac{z}{k}\right)(k+z)^{n-1} + \frac{1}{n!}(k+z)^{n-1} - \sum_{i=1}^n \frac{H_{n} -H_{n-i}}{(i-1)!( n-i) !} k^{n-i}z^{i-1}$$

Then, what remains to be shown is $$\frac{1}{n!}(k+z)^{n-1} - \sum_{i=1}^n \frac{H_{n} -H_{n-i}}{(i-1)!( n-i) !} k^{n-i}z^{i-1} = -\sum ^{n-1}_{i=1}\frac{H_{n-1} -H_{n-1-i}}{i!( n-1-i) !} k^{n-1-i} z^{i}$$

Expanding $(k+z)^{n-1}$ by the binomial expansion and setting the coefficients of $z^i$ equal to each other yields $$\frac{1}{n!}\binom{n-1}{i}k^{n-1-i} - \frac{H_{n} -H_{n-i-1}}{i!( n-i-1) !} k^{n-i-1}=-\frac{H_{n-1} -H_{n-1-i}}{i!( n-1-i) !} k^{n-1-i}$$

Subtracting $\frac{H_{n-1-i}}{i!(n-1-i)!}k^{n-1-i}$, dividing by $k^{n-1-i}$, then multiplying by $i!(n-1-i)!$ on both sides makes it $$\frac{(n-1)!}{n!} - H_n = -H_{n-1}$$

This simplifies to $$\frac{1}{n}-H_n = -H_{n-1} \to H_n = H_{n-1} + \frac{1}{n}$$

which is exactly the recursive definition of the harmonic numbers. Therefore, your conjecture is true.

Answer 3

I realized that showing how I came up with my conjecture could probably provide the basis of a proof. I'm not sure if I should post this as a separate answer or as part of the question, but since it kind of answers the question I'm posting as an answer. I will not, however, accept this.

I already knew the answer would be in the form $$\left(\frac d{dz}\right)^{-n}\ln\left(1+\frac zk\right)=\frac{\ln\left(1+\frac zk\right)-H_n}{n!}(k+z)^n+P(z,k,n),$$ for some polynomial $P$, thanks in part to the nth integral of $\ln(z)$: \begin{aligned} \left(\frac d{dz}\right)^{-n}\ln\left(1+\frac zk\right)&=\left(\frac d{d(1+z/k)}\frac{d(1+z/k)}{dz}\right)^{-n}\ln\left(1+\frac zk\right)\\ &\rightsquigarrow\left(\frac d{d(1+z/k)}\right)^{-n}\ln\left(1+\frac zk\right)\left(\frac{d(1+z/k)}{dz}\right)^{-n}\\ &=\frac{\ln\left(1+\frac zk\right)-H_n}{n!}\left(1+\frac zk\right)^n\left(\frac1k\right)^{-n}\\ &=\frac{\ln\left(1+\frac zk\right)-H_n}{n!}(k+z)^n. \end{aligned} I made a table of $P$ to hopefully find a pattern \begin{matrix} n & P(z,k,n)\\ 0 & 0\\ 1 & k\\ 2 & kz+\frac{3k^2}4\\ 3 & \frac{kz^2}2+\frac{3k^2z}4+\frac{11k^3}{36}\\ 4 & \frac{kz^3}3+\frac{3k^2z^2}8+\frac{11k^3z}{36}+\frac{25k^4}{288}\\ \dots & \dots \end{matrix} and indeed several nice patterns appear. The table shows that we can expect $P$ to be in the form $$P(z,k,n)=\sum_{i=0}^n\frac{H_i}{i!(n-i)!}z^{n-i}k^i,$$ which then leads to \begin{aligned} \left(\frac d{dz}\right)^{-n}\ln\left(1+\frac zk\right)&=\frac{\ln\left(1+\frac zk\right)-H_n}{n!}(k+z)^n+\sum_{i=0}^n\frac{H_i}{i!(n-i)!}z^{n-i}k^i\\ &=\frac1{n!}\ln\left(1+\frac zk\right)(k+z)^n-\sum_{i=0}^n\frac{H_n-H_i}{i!(n-i)!}z^{n-i}k^i. \end{aligned} Thus, all that really has to be done is prove this form of $P$.