In this post we defined the following
$$\mathscr{H}(p,q) = \sum_{k\geq 1}\frac{\mathscr{C}(p,k)}{k^q}$$
Hence we have the following integral representation
$$\mathscr{H}(p,q)= \int^1_0 \frac{\mathrm{Li}_p(x) \, \mathrm{Li}_q(x)}{x}\, dx$$
\begin{align}
\mathscr{H}(p,q)=\int^1_0 \frac{\mathrm{Li}_p(x)\,\, \mathrm{Li}_q(x)\, }{x}\, dx&= \sum_{n=1}^{p-1}(-1)^{n-1}\zeta(p-n+1)\zeta(q+n) -\frac{1}{2}\sum_{n=1}^{{p+q}-2}(-1)^{p-1}\zeta(n+1)\zeta({p+q}-n)\\ &+(-1)^{p-1}\left(1+\frac{{p+q}}{2} \right)\zeta({p+q}+1)\end{align}
Now for the special case \( p=q\) we have
\begin{align}
\mathscr{H}(q,q)=\int^1_0 \frac{ \mathrm{Li}_q(x) ^2 \, }{x}\, dx&= \sum_{n=1}^{q-1}(-1)^{n-1}\zeta(q-n+1)\zeta(q+n) -\frac{1}{2}\sum_{n=1}^{{2q}-2}(-1)^{q-1}\zeta(n+1)\zeta({2q}-n)\\ &+(-1)^{q-1}\left(1+q \right)\zeta(2q+1)\end{align}
We have for the special case \(q=2\)
$$\mathscr{H}(2,2)=\int^1_0 \frac{ \mathrm{Li}_2(x) ^2 \, }{x}\, dx=2\zeta(3)\zeta(2)-3\zeta(5)$$
$$\mathscr{H}(p,q) = \sum_{k\geq 1}\frac{\mathscr{C}(p,k)}{k^q}$$
Hence we have the following integral representation
$$\mathscr{H}(p,q)= \int^1_0 \frac{\mathrm{Li}_p(x) \, \mathrm{Li}_q(x)}{x}\, dx$$
\begin{align}
\mathscr{H}(p,q)=\int^1_0 \frac{\mathrm{Li}_p(x)\,\, \mathrm{Li}_q(x)\, }{x}\, dx&= \sum_{n=1}^{p-1}(-1)^{n-1}\zeta(p-n+1)\zeta(q+n) -\frac{1}{2}\sum_{n=1}^{{p+q}-2}(-1)^{p-1}\zeta(n+1)\zeta({p+q}-n)\\ &+(-1)^{p-1}\left(1+\frac{{p+q}}{2} \right)\zeta({p+q}+1)\end{align}
Now for the special case \( p=q\) we have
\begin{align}
\mathscr{H}(q,q)=\int^1_0 \frac{ \mathrm{Li}_q(x) ^2 \, }{x}\, dx&= \sum_{n=1}^{q-1}(-1)^{n-1}\zeta(q-n+1)\zeta(q+n) -\frac{1}{2}\sum_{n=1}^{{2q}-2}(-1)^{q-1}\zeta(n+1)\zeta({2q}-n)\\ &+(-1)^{q-1}\left(1+q \right)\zeta(2q+1)\end{align}
We have for the special case \(q=2\)
$$\mathscr{H}(2,2)=\int^1_0 \frac{ \mathrm{Li}_2(x) ^2 \, }{x}\, dx=2\zeta(3)\zeta(2)-3\zeta(5)$$
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