TITLE: Closed-form of $\int_0^1 \operatorname{Li}_p(x) \, dx$
QUESTION [11 upvotes]: 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?

REPLY [4 votes]: 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
$$
\begin{align}
\int_0^1\mathrm{Li}_p(x)\,\mathrm{d}x
&=\sum_{k=0}^{p-2}(-1)^k\zeta(p-k)+(-1)^{p-1}\sum_{k=1}^\infty\frac1{k(k+1)}\\
&=\sum_{k=0}^{p-2}(-1)^k\zeta(p-k)+(-1)^{p-1}
\end{align}
$$