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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. :)

Alejandro Jimenez Tellado
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    A small observation, assuming your calculations up to this point are right. You could use partial fractions on the series to get $I(n,2) = n \sum_{m\geq 1} \left( \frac{1}{nm-1} - \frac{1}{nm}\right) = n\sum_{j \geq 1} \frac{b_j}{j}$ where $b_j= -1$ if $j=0 mod n$, $b_j= 1$ if $j=-1 mod n$, and $b_j= 0$ else. You can express $b_j$ as a combination of powers of roots of unity. It will be tedious but I think it will allow you to get to a closed form. – Lee Fisher Jul 10 '23 at 19:38
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    The last sum seems to be $-\psi\left(\frac{n - 1}{n}\right) - \gamma$. – Travis Willse Jul 10 '23 at 21:28
  • Cf. https://math.stackexchange.com/questions/2342476/an-integral-related-to-the-digamma-function – Travis Willse Jul 10 '23 at 21:33
  • @TravisWillse I don't think that link is relevant as the real part of the power in the denominator must be greater than 1 which is not the case here, after the appropriate substitution to get the integral to have the same bounds. – Alejandro Jimenez Tellado Jul 10 '23 at 21:36
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    To be clear, my two previous comments are unrelated. The link only points to a similar family of integrals. – Travis Willse Jul 10 '23 at 21:38
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    (Also, is the argument of $\ln$ in the first integral missing an $x$ term?) – Travis Willse Jul 10 '23 at 21:38
  • Just double-checked and it does not. I did a u-sub for 1/x and then did IBP. – Alejandro Jimenez Tellado Jul 10 '23 at 21:42
  • @AlejandroJimenezTellado I mean the very first integral in the post, the one with value $\frac{\pi^2}{7}$. – Travis Willse Jul 10 '23 at 21:57
  • Oh no it's certainly correct, you use partial geometric series and the results I listed underneath. – Alejandro Jimenez Tellado Jul 10 '23 at 21:58
  • Notice that your summation is the same as my result – Claude Leibovici Jul 11 '23 at 04:41

3 Answers3

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

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

Claude Leibovici
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    I also did the same but I am getting $$ I(n,m) = \frac{1}{m-1} \mathrm{H}_{\frac{1-m}{n}} $$ I wanted to write $I(n,m)$ in terms of $\psi$ function but $\psi(o)$ is undefined I guess? – Lucky Chouhan Jul 11 '23 at 05:30
  • Awesome stuff, your solutions always impress me. Though I do feel very dumb in hindsight because I knew that $\sum{1/(n(n+m))}=H_m/m$ and just forgot that $H_{(-1/n)}$ was a thing. But with that being said do you think there is any way to proceed forward with $I(n,3)$ for instance? I will look at it myself of course, but just wondering if you had any immediate ideas. Thank you again :) – Alejandro Jimenez Tellado Jul 11 '23 at 13:06
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    @AlejandroJimenezTellado. I gave the general formula. These are just generalized harmonic numbers. If you consider that the result is $a,H_{-\frac 1a}$ you can make nice series expansion for large $a$ – Claude Leibovici Jul 11 '23 at 13:24
  • Omg I'm so sorry it's too early in the morning for me, I'm going back to bed, thank you again. – Alejandro Jimenez Tellado Jul 11 '23 at 13:33
  • @ClaudeLeibovici. I think you have the wrong solution for $I(m,n)$. Please if you can, see my answer and confirm. – bob Aug 09 '23 at 18:17
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    @LuckyChouhan. I also got the same result. To re-write in terms of the $\psi$ function you can use the relation $$H_z=\psi(z+1)+\gamma$$ – bob Aug 10 '23 at 09:32
  • Yeah! @bob $$ I(n.m) = \frac{1}{m-1} \left( \psi \left( \frac{1-m+n}{n} \right) - \gamma \right) $$ – Lucky Chouhan Aug 10 '23 at 11:52
  • Glad to see you back @ClaudeLeibovici. – bob Aug 13 '23 at 05:16
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    @bob. Thanks ! I was really sorry to not fix it earlier but I was unable to post earlier. – Claude Leibovici Aug 13 '23 at 05:19
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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 $I(2,5)$, $$\begin{align*}\int_0^1\frac{\log\left(1-x^{5}\right)}{x^{2}}\ dx&=-0.38779290180\dots \\ H_{-1/5}&=-0.38779290180\dots\end{align*}$$

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

$E$: Euler's definition of Harmonic number

$IBP$: Integration by parts

Bob Dobbs
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