Closed-form of $\int_0^1 \operatorname{Li}_p(x) \, dx$

Question

While I've studied integrals involving polylogarithm functions I've observed that

$$\int_0^1 \operatorname{Li}_p(x) \, dx \stackrel{?}{=} \sum_{k=2}^p(-1)^{p+k}\zeta(k)+(-1)^{p+1},\tag{1}$$

for any integer $p\geq2$. Here $\zeta$ is the Riemann zeta function.

After that I have three questions.

• $1^\text{st}$ Question. Is $(1)$ true? If it is, how could we prove it?
• $2^\text{nd}$ Question. If it's a well-known result could you give any reference?
• $3^\text{rd}$ Question. I think there is also a similar closed-form of $\int_0^b \operatorname{Li}_p(x) \, dx$, for any integer $b \geq 1$. What is the closed-form of this integral?

Answer 1

A simple integration by parts gives us a very simple recurrence relation:

$$\begin{align} \int_{0}^{1}\operatorname{Li}_{p}{\left(x\right)}\,\mathrm{d}x &=\left[x\operatorname{Li}_{p}{\left(x\right)}\right]_{0}^{1}-\int_{0}^{1}x\frac{d}{dx}\left(\operatorname{Li}_{p}{\left(x\right)}\right)\,\mathrm{d}x\\ &=\operatorname{Li}_{p}{\left(1\right)}-\int_{0}^{1}x\cdot\frac{\operatorname{Li}_{p-1}{\left(x\right)}}{x}\,\mathrm{d}x\\ &=\zeta{(p)}-\int_{0}^{1}\operatorname{Li}_{p-1}{\left(x\right)}\,\mathrm{d}x\\ \end{align}$$

From there, the desired statement can be proven via induction in the usual manner.

Answer 2

\begin{align} \int^1_0{\rm Li}_q(x)\ {\rm d}x =&\sum^\infty_{n=1}\frac{1}{n^q(n+1)}\tag1\\ =&\sum^\infty_{n=1}\left[\frac{(-1)^q}{n+1}+\sum^q_{j=1}\frac{1}{n^j}\operatorname*{Res}_{n=0}\frac{1}{n^{q-j+1}(n+1)}\right]\tag2\\ =&\sum^\infty_{n=1}\left[\frac{(-1)^q}{n+1}+\sum^q_{j=1}\frac{1}{n^j}\cdot\frac{1}{(q-j)!}\cdot\frac{\partial^{q-j}}{\partial n^{q-j}}\frac{1}{n+1}\Bigg{|}_{n=0}\right]\tag3\\ =&\sum^\infty_{n=1}\left[(-1)^q\left[\frac{1}{n+1}-\frac{1}{n}\right]+\sum^q_{j=2}\frac{(-1)^{q-j}}{n^j}\right]\tag4\\ =&(-1)^{q+1}+\sum^q_{j=2}\sum^\infty_{n=1}\frac{(-1)^{q-j}}{n^j}\tag5\\ =&(-1)^{q+1}+\sum^q_{j=2}(-1)^{q-j}\zeta(j) \end{align}

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Explanation:
$(1): \text{Expand the polylogarithm and integrate term by term}$
$(2): \text{Partial fractions}$
$(3): \text{Compute the residue}$
$(4): \text{Rearrange terms}$
$(5): \text{The first sum telescopes; Swap the order of summation for the second}$

Answer 3

In several places, we will use $\frac1{k(k+1)}=\frac1k-\frac1{k+1}$. $$ \begin{align} \int_0^1\sum_{k=1}^\infty\frac{x^k}{k^p}\mathrm{d}x &=\color{#C00000}{\sum_{k=1}^\infty\frac1{k^p(k+1)}}\\ &=\sum_{k=1}^\infty\left(\frac1{k^p}-\frac1{k^{p-1}(k+1)}\right)\\ &=\color{#C00000}{\zeta(p)-\sum_{k=1}^\infty\frac1{k^{p-1}(k+1)}}\\ \end{align} $$ Iterating the recursion in red, we get $$ \begin{align} \int_0^1\mathrm{Li}_p(x)\,\mathrm{d}x &=\sum_{k=0}^{p-2}(-1)^k\zeta(p-k)+(-1)^{p-1}\sum_{k=1}^\infty\frac1{k(k+1)}\\ &=\sum_{k=0}^{p-2}(-1)^k\zeta(p-k)+(-1)^{p-1} \end{align} $$