Evaluating a Logarithmic Integral

Question

For everything on this post $n$ and $m$ are positive integers.

The other day I found the following integral on the popular post "Integral Milking" and decided to give it a go.

$$\large\int_{0}^{1} {\frac{\ln{(1+x^2+x^3+x^4+x^5+x^6)}}{x}}dx=\frac{{\pi}^2}{7}$$

After some simple manipulations, I eventually had

$$\large\int_{0}^{1} {\frac{\ln{(1+x+...+x^n)}}{x}}dx=\frac{\pi^2}{6}\frac{n}{n+1}.$$

But more interestingly in order to get that result I found

$$\large\int_{0}^{1} {\frac{\ln{(1-x^n)}}{x}}dx=-\frac{\pi^2}{6n},$$

which sparked a number of thoughts relating to the further generalization of the following,

$$I(n,m)=\large\int_{0}^{1} {\frac{\ln{(1-x^n)}}{x^m}}dx,\hspace{5pt}n\geq{m}.$$

Surprisingly I am already stuck on $m=2$ and have only made a small bit of progress in the following two ways.

$$I(n,2)=\large\int_{1}^{\infty} {\frac{1}{x^n-1}}dx+\left(x\ln{\left(1-\frac{1}{x^n}\right)}\right)\bigg|_1^{\infty}$$

$$I(n,2)=\large\sum_{m=1}^{\infty} {\frac{1}{m(nm-1)}}$$

I have a strong suspicion that these have both been asked before, but I have been struggling to find them, and if indeed there is already a post satisfying my intrigue I will delete this one. Regardless it seems there is some Digamma and/or a nice Hypergeometric Series result afoot that I have yet to consider. That or the last integral is simply trivial and I am just dumb. :)

Answer 1

Expanding the logarithm by its Taylor series, $$\int_0^1\frac{\log(1-x^n)}{x^m}\ dx=-\int_0^1\sum_{k=1}^\infty\frac{x^{nk-m}}{k}\ dx=-\sum_{k=1}^\infty\frac{1}{k}\int_0^1x^{nk-m}\ dx=-\sum_{k=1}^\infty\frac{1}{k(nk-m+1)}$$ and as, $$\begin{align*}-\sum_{k=1}^\infty\frac{1}{k(nk-m+1)}&=-\sum_{k=1}^\infty\frac{1/n}{k(k+(1-m)/n)} \\ &=\frac{1}{m-1}\sum_{k=1}^\infty\frac{\color{teal}{(1-m)/n}}{k(k+\color{teal}{(1-m)/n})}\end{align*}$$ by the formula, $$H_z=\psi(z+1)+\gamma=\sum_{k=1}^\infty\frac{z}{k(k+z)},\quad z\neq -1,-2,-3,\dots$$ taking $z=(1-m)/n$ and multiplying by $1/(m-1)$ yields, $$\frac{1}{m-1}\sum_{k=1}^\infty\frac{(1-m)/n}{k(k+(1-m)/n)}=\frac{{1}}{m-1}H_{\frac{1-m}{n}}$$ thus the integral is actually, $$\boxed{I(m,n)=\frac{1}{m-1}H_{\frac{1-m}{n}}.}$$ Numerical check with $I(2,5)$, $$\begin{align*}\int_0^1\frac{\log\left(1-x^{5}\right)}{x^{2}}\ dx&=-0.38779290180\dots \\ H_{-1/5}&=-0.38779290180\dots\end{align*}$$

Answer 2

$$I(n,m)=\int_0^1 {\frac{\log{(1-x^n)}}{x^m}}\,dx=\frac 1n\int_0^1 \log (1-t)\, t^{-\frac{m+n-1}{n}}\,dt$$

Provided that $a>-2$ $$\int_0^1 \log(1-t)\, t^a\,dt=\frac{H_{a+1}}{a+1}$$ where $H_k$ is the harmonic number. So $$I(n,m)=\frac 1 {m-1}\,H_{\frac{1-m}{n}}$$ which gives the nice $$I(n,2)=-H_{-\frac{1}{n}}$$

Using one integration by parts, the antiderivative is given in terms of the Gaussian hypergeometric function $$(a+1)(a+2)\int \log(1-t)\, t^a\,dt=t^{a+2} \, _2F_1(1,a+2;a+3;t)+$$ $$(a+2) t^{a+1} \log (1-t)$$

In the previous edit, there was a mistake underlined by @bob

Answer 3

$$\int_0^1\frac{\ln{(1-x^n)}}{x^m}dx\\\stackrel{t=x^n}=\frac1n\int_0^1\ln(1-t)t^{\frac{1-m-n}{n}}dt\\ \stackrel{IBP}=\frac1{m-1}\int_0^1\frac{t^{\frac{1-m}{n}}-1}{t-1}dt\\ \stackrel{E}=\frac1{m-1}H_{\frac{1-m}{n}}$$ where $m<n+1$.

$E$: Euler's definition of Harmonic number

$IBP$: Integration by parts