Closed forms of the integral $ \int_0^1 \frac{\mathrm{Li}_n(x)}{(1+x)^n} d x $

Question

(This is related to this question).

How would one find the closed forms the integral $$ \int_0^1 \frac{\mathrm{Li}_n(x)}{(1+x)^n} d x? $$

I tried using Nielsen Generalized Polylogarithm as mentioned in the linked post, but it doesn't produce anything. Another approach I tried was IBP, which seemed a bit too tedious. Does anyone have a closed form of this function? Thank you!

Answer 1

For integer $n, m$ such $1 \leqslant m\leqslant n$ let $$ \mathfrak{I}(n,m) = \int\limits_{0}^{1} \dfrac{\operatorname{Li}_{n}(x)}{(1+x)^{m}}\,\mathrm{d}x. \label{i_n_m}\tag{1} $$ Also, from the linked answer we know the closed form of $\mathfrak{I}(n,1)$ for $n \geqslant 1$.

By induction, it can be proved that $$ \dfrac{1}{x(1+x)^{m}} = \dfrac{1}{x} - \sum\limits_{k=1}^{m}\dfrac{1}{(1+x)^{k}}, \quad m \geqslant 0 \label{frac_expand}\tag{2} $$

Let $m \geqslant 2$. Then

\begin{align} \mathfrak{I}(n,m) &\overset{\text{IBP}}{=} -\dfrac{1}{(m-1)2^{m-1}}\zeta(n) +\dfrac{1}{m-1}\int\limits_{0}^{1} \dfrac{\operatorname{Li}_{n-1}(x)}{x(1+x)^{m-1}}\,\mathrm{d}x \\ &\overset{\eqref{frac_expand}}{=} \dfrac{1}{m-1}\left(1-\dfrac{1}{2^{m-1}}\right)\zeta(n)-\dfrac{1}{m-1}\sum\limits_{k=1}^{m-1}\mathfrak{I}(n-1,k) \label{receq}\tag{3} \end{align}

From the last relation we can conclude that $\mathfrak{I}(n,m)$ can be expressed as linear combination of \begin{align} \mathfrak{I}(p, 1)&,\quad 1 \leqslant p \leqslant m-1, \label{i_dep}\tag{4}\\ \zeta(n-k)&,\quad 0 \leqslant k \leqslant m-2, \end{align}

and we want to find the coefficients in front of them.

Pretend that we use formula \eqref{receq} one more time for every entry on the right with second parameter $k \geqslant 2$. And we repeat that operation $p \geqslant 1$ times starting from $\mathfrak{I}(n,m)$. Observe that now $\mathfrak{I}(n,m)$ will depend on $\zeta, \mathfrak{I}(\cdot, 1)$ and $$\mathfrak{I}(n-p, m-k),\ \ p \leqslant k \leqslant m-1. \label{expand_p} \tag{5}$$

We denote as $\tau_{n,m}(p,k)$ the coefficient in front of $\mathfrak{I}(n-p, m-k)$ in that representation and calculate it in a combinatorial way.

We start from $\mathfrak{I}(n, i_{p}),\ \ i_p = m$, expand it with formula \eqref{receq}, select a $\mathfrak{I}(n-1, i_{p-1}),\ \ m-k < i_{p-1} < i_{p}$ and repeat the process of expanding. Then we will obtain a "partial" coefficient $$ (-1)^{p}\prod\limits_{j=1}^{p} \dfrac{1}{i_{j}-1} = \dfrac{(-1)^{p}}{m-1}\prod\limits_{j=1}^{p-1} \dfrac{1}{i_{j}-1}. $$ To get total coefficient, we need to sum all "partial" coefficients: \begin{align} \tau_{n,m}(p,k) &= \dfrac{(-1)^{p}}{m-1}\sum\limits_{\substack{i_{q} < i_{q+1}\\m-k < i_{q} < m }}\prod\limits_{j=1}^{p-1} \dfrac{1}{i_{j}-1} \\ &= \dfrac{(-1)^{p}}{m-1}\sum\limits_{\substack{i_{q} < i_{q+1}\\0 < i_{q} < k}}\prod\limits_{j=1}^{p-1} \dfrac{1}{i_{j}+m-k-1} \\ &= \dfrac{(-1)^{p}}{(m-1)}\sum\limits_{\substack{i_{q} < i_{q+1}\\0 < i_{q} < k}}\prod\limits_{j=1}^{k-1} \dfrac{1}{(j+m-k-1)}\prod\limits_{j \neq i_{q}} (j+m-k-1) \\ &= \dfrac{(-1)^{p}}{(m-1)\ldots(m-k)}\sum\limits_{\substack{i_{q} < i_{q+1}\\0 < i_{q} < k}}\prod\limits_{j=1}^{k-p} (i_{j}+m-k-1) \\ &= \dfrac{(-1)^{p}}{(m-1)\ldots(m-k)}\sum\limits_{\substack{i_{q} < i_{q+1}\\0 < i_{q} < k}}\prod\limits_{j=1}^{k-p} (i_{j}+m-k-1) \\ &= \dfrac{(-1)^{k}}{(m-1)\ldots(m-k)}[x^{p-1}](x-m+2)_{\overline{k-1}}. \label{coeff}\tag{6} \end{align}

At the last step we used a general Vieta formula. Also, $[x^{p}]$ refers to a coefficient in front of the $x^{p}$ in the expression following it.

When $k=m-1$, from the definition of unsigned Stirling numbers of the first kind we get $$ \tau_{n,m}(p,m-1) = \dfrac{(-1)^{p}}{(m-1)!}\left[{m-1 \atop p}\right] $$

Now we need to note two things:

• After $p$-th operation a coefficient in front of $\mathfrak{I}(n-i, 1),\ \ 1 \leqslant i \leqslant p$ doesn't change. It follows directly from the fact that in \eqref{expand_p} values with $p \leqslant k \leqslant m-2$ have no $\mathfrak{I}(n-i, 1)$ due \eqref{i_dep}.
• $\zeta(k)$ can only occur after applying \eqref{receq} to a $\mathfrak{I}(k, m),\ \ m \geqslant 2$.

From the first note it follows that $\tau_{n,m}(p,m-1)$ is the coefficient in front of a $\mathfrak{I}(n-p, 1)$.

The second note implies the formula for a coefficient $\lambda_{n,m}(p)$ in front of the $\zeta(n-p)$: \begin{align} \lambda_{n,m}(p) &= \sum\limits_{k=p}^{m-2}\dfrac{\tau_{n,m}(p,k)}{m-k-1}\left(1-\dfrac{1}{2^{m-k-1}}\right) \\ &= [x^{p-1}]\sum\limits_{k=p}^{m-2}\dfrac{(-1)^{k}(x-m+2)_{\overline{k-1}}}{(m-k-1)\ldots(m-1)}\left(1-\dfrac{1}{2^{m-k-1}}\right) \\ &= \dfrac{[x^{p-1}]}{(m-1)!}\sum\limits_{k=p}^{m-2}(-1)^{k}(m-2-k)!\left(1-\dfrac{1}{2^{m-k-1}}\right)\sum\limits_{\alpha=0}^{k-1}\left[{k-1 \atop \alpha}\right](x-m+2)^{\alpha} \\ &= \dfrac{[x^{p-1}]}{(m-1)!}\sum\limits_{k=p}^{m-2}(-1)^{k}(m-2-k)!\left(1-\dfrac{1}{2^{m-k-1}}\right)\sum\limits_{\alpha=0}^{k-1}\left[{k-1 \atop \alpha}\right]\sum\limits_{\gamma = 0}^{\alpha}\binom{\alpha}{\gamma}x^{\gamma}(2-m)^{\alpha-\gamma} \\ &=\dfrac{1}{(m-1)!}\sum\limits_{k=p}^{m-2}(-1)^{k}(m-2-k)!\left(1-\dfrac{1}{2^{m-k-1}}\right)\sum\limits_{\alpha=p}^{k}\left[{k-1 \atop \alpha-1}\right]\binom{\alpha-1}{p-1}(2-m)^{\alpha-p} \\ &=\dfrac{1}{(m-1)!}\sum\limits_{\alpha=p}^{m-2}\binom{\alpha-1}{p-1}(2-m)^{\alpha-p}\sum\limits_{k=\alpha}^{m-2}\left[{k-1 \atop \alpha-1}\right](-1)^{k}(m-2-k)!\left(1-\dfrac{1}{2^{m-k-1}}\right) \end{align}

So, for $2 \leqslant m \leqslant n$ we have \begin{align} \mathfrak{I}(n,m) &= \dfrac{1}{(m-1)!}\sum\limits_{p=1}^{m-1} (-1)^{p}\left[{m-1 \atop p}\right]\mathfrak{I}(n-p,1) \\ &+ \dfrac{1}{m-1}\left(1-\dfrac{1}{2^{m-1}}\right)\zeta(n) + \sum\limits_{p=1}^{m-2}\lambda_{n,m}(p)\zeta(n-p), \end{align}

where $\lambda_{n,m}(p)$ is defined above.