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I'm unsure if this has been discovered already, but it's heavily related to my current research, particularly to this question of mine (this conjecture was also originally posted at the beginning of my most recent question). My conjecture is the following:

Let $f(x)$ be continuous on the open interval $a\lt x\lt b$. If $\int_a^b f(x)\,dx=0$, then $$\sum_{k=1}^\infty\int_a^b\mathrm{Li}_k(f(x))\,dx=-\int_a^bf(x)\ln(1-f(x))\,dx,\tag{*}$$ where $\mathrm{Li}_k(z)$ represents the polylogarithm of order $k$.

I've made an assumption along the way during my proof, and I need help in showing that it's justified.

We start with the series definition for polylogarithms to obtain $$\sum_{k=1}^\infty\int_a^b\mathrm{Li}_k(f(x))\,dx=\sum_{k=1}^\infty\sum_{n=1}^\infty\int_a^b\frac{(f(x))^n}{n^k}dx.\tag{1}$$ Since we are given the initial condition $\int_a^b f(x)\,dx=0$, the value of the series at $n=1$ is also $0$. Therefore, we have $$\begin{align} \sum_{k=1}^\infty\int_a^b\mathrm{Li}_k(f(x))\,dx&=\color{blue}{\sum_{k=1}^\infty\sum_{n=2}^\infty\int_a^b\frac{(f(x))^n}{n^k}dx}\tag{2}\\&=\color{red}{\sum_{n=2}^\infty\sum_{k=1}^\infty\int_a^b\frac{(f(x))^n}{n^k}dx}.\tag{3}\\ \end{align}$$ Noting that the inner sum is a geometric series, we get $$\sum_{n=2}^\infty\sum_{k=1}^\infty\int_a^b\frac{(f(x))^n}{n^k}dx=\sum_{n=2}^\infty\int_a^b\frac{(f(x))^n}{n-1}dx.\tag{4}$$ Finally, using the Mercator series gives us $$\sum_{n=2}^\infty\int_a^b\frac{(f(x))^n}{n-1}dx=-\int_a^b f(x)\ln(1-f(x))\,dx,\tag{5}$$ as desired.

Between equations 2 and 3, I assumed that swapping the order of summation was justified. However, this would require showing that the series converges for all $f(x)$ satisfying the initial conditions (see here). The conjecture holds for $f(x)=\ln{x}$ with $0\lt x\lt e$ (see here), $f(x)=\cos{x}$ for $0\lt x\lt \pi$, and appears to hold numerically for $f(x)=\frac{\ln{x}}{x^2+1}$ for $0\lt x\lt \infty$. This leads to another problem: the series definition for $\mathrm{Li}_s(z)$ only holds for $|z|\lt1$. However, for some reason, this conjecture appears to avoid that restriction entirely. It also appears to hold even if $f(x)\gt1$, but the resulting value would be complex. Needless to say, I'm completely lost.

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  • Notice that analytic continuation can come into play – Simply Beautiful Art Nov 13 '16 at 16:53
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    If $$\sum_{n=1}^\infty f(n):=\lim_{N\to\infty}\sum_{n=1}^Nf(n)$$ then we need to prove $$\sum_{n=1}^\infty\sum_{k=1}^\infty f(n,k)=\lim_{N\to\infty}\lim_{K\to\infty}\sum_{n=1}^N\sum_{k=1}^Kf(n,k)=\lim_{K\to\infty}\lim_{N\to\infty}\sum_{n=1}^N\sum_{k=1}^Kf(n,k)$$ where we have double limit? Show taking limit with $K$ first or $N$ first that it comes out the same. – Simply Beautiful Art Nov 13 '16 at 16:57
  • @SimpleArt I haven't taken a formal course in real analysis, so I'm a bit confused. The equivalence holds trivially if $|f(x)|\leq1$ is true for $a\lt x\lt b$, but otherwise my approach seems invalid. Do I need to scrap the proof entirely and use another method? – teadawg1337 Nov 14 '16 at 15:44
  • No no, I mean that to have two infinite sums, you consider the partial sums, by definition, then rearrange. And as long as you derive an analytic solution for a dense set of values, by analytic continuation, it will hold everywhere the solution exists. – Simply Beautiful Art Nov 14 '16 at 17:40
  • Analytic continuation is a powerful tool. – Simply Beautiful Art Nov 14 '16 at 17:55
  • @SimpleArt So proving that both of the partial sums converge would be sufficient? – teadawg1337 Nov 15 '16 at 13:28
  • Yes, because switching the finite sums are allowed. – Simply Beautiful Art Nov 15 '16 at 14:30

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