By using
$$\int_0^\infty\frac{\ln^{2n}(x)}{1+x^2}dx=|E_{2n}|\left(\frac{\pi}{2}\right)^{2n+1}\tag{a}$$
and
$$\text{Li}_{a}(-z)+(-1)^a\text{Li}_{a}(-1/z)=-2\sum_{k=0}^{\lfloor{a/2}\rfloor }\frac{\eta(2k)}{(a-2k)!}\ln^{a-2k}(z)\tag{b}$$
I managed to find:
$$\int_0^\infty\frac{\ln^{2n}(x)}{1+yx^2}dx=\frac{\left(\frac{\pi}{2}\right)^{2n+1}}{\sqrt{y}}\sum_{k=0}^n\binom{2n}{2k}|E_{2n-2k}|\pi^{-2k}\ln^{2k}(y)\tag{c}$$
$$\int_0^\infty\frac{\ln^{2n-1}(x)}{1+yx^2}dx=\frac{-\left(\frac{\pi}{2}\right)^{2n-1}}{2\sqrt{y}}\sum_{k=0}^{n-1}\binom{2n-1}{2k+1}|E_{2n-2k-2}|\pi^{-2k}\ln^{2k+1}(y)\tag{d}$$
$$\int_0^\infty\frac{\ln^{2n}(x)\text{Li}_{2a+1}(-x^2)}{1+x^2}dx=|E_{2n}|\left(\frac{\pi}{2}\right)^{2n+1}\zeta(2a+1)$$ $$-\frac{\left(\frac{\pi}{2}\right)^{2n+1}}{(2a)!}\sum_{k=0}^n \binom{2n}{2k}|E_{2n-2k}|\pi^{-2k}(2a+2k)!(2^{2k+2a+1}-1)\zeta(2k+2a+1)\tag{e}$$
$$\int_0^\infty\frac{\ln^{2n-1}(x)\text{Li}_{2a}(-x^2)}{1+x^2}dx=\frac{-\left(\frac{\pi}{2}\right)^{2n-1}}{2(2a-1)!}*$$ $$\sum_{k=0}^{n-1} \binom{2n-1}{2k+1}|E_{2n-2k-2}|\pi^{-2k}(2a+2k)!(2^{2k+2a+1}-1)\zeta(2k+2a+1)\tag{f}$$
$$\int_0^1\frac{\ln^{2n}(x)\text{Li}_{2a+1}(-x^2)}{1+x^2}dx=\frac12|E_{2n}|\left(\frac{\pi}{2}\right)^{2n+1}\zeta(2a+1)$$ $$-\frac{\left(\frac{\pi}{2}\right)^{2n+1}}{2(2a)!}\sum_{k=0}^n \binom{2n}{2k}|E_{2n-2k}|\pi^{-2k}(2a+2k)!(2^{2k+2a+1}-1)\zeta(2k+2a+1)$$ $$+2\sum_{k=0}^a\frac{(2k+2n+1)!}{(2k+1)!}4^k \eta(2a-2k)\beta(2k+2n+2)\tag{g}$$
$$\int_0^1\frac{\ln^{2n-1}(x)\text{Li}_{2a}(-x^2)}{1+x^2}dx=$$ $$\frac{-\left(\frac{\pi}{2}\right)^{2n-1}}{4(2a-1)!}\sum_{k=0}^{n-1} \binom{2n-1}{2k+1}|E_{2n-2k-2}|\pi^{-2k}(2a+2k)!(2^{2k+2a+1}-1)\zeta(2k+2a+1)$$ $$+\sum_{k=0}^a\frac{(2k+2n-1)!}{(2k)!}4^k \eta(2a-2k)\beta(2k+2n)\tag{h}$$
$$\sum_{k=1}^\infty\frac{(-1)^k H^{(2a+1)}_k}{(2k+1)^{2n+1}}=\frac1{2(2n)!}|E_{2n}|\left(\frac{\pi}{2}\right)^{2n+1}\zeta(2a+1)$$ $$-\frac{\left(\frac{\pi}{2}\right)^{2n+1}}{2(2n)!(2a)!}\sum_{k=0}^n \binom{2n}{2k}|E_{2n-2k}|\pi^{-2k}(2a+2k)!(2^{2k+2a+1}-1)\zeta(2k+2a+1)$$ $$+\frac{2}{(2n)!}\sum_{k=0}^a\frac{(2k+2n+1)!}{(2k+1)!}4^k \eta(2a-2k)\beta(2k+2n+2)\tag{i}$$
$$\sum_{k=1}^\infty\frac{(-1)^k H^{(2a)}_k}{(2k+1)^{2n}}=\frac{\left(\frac{\pi}{2}\right)^{2n-1}}{4(2n-1)!(2a-1)!}*$$ $$\sum_{k=0}^{n-1} \binom{2n-1}{2k+1}|E_{2n-2k-2}|\pi^{-2k}(2a+2k)!(2^{2k+2a+1}-1)\zeta(2k+2a+1)$$ $$-\frac{1}{(2n-1)!}\sum_{k=0}^a\frac{(2k+2n-1)!}{(2k)!}4^k \eta(2a-2k)\beta(2k+2n)\tag{j}$$
Question: Are the results of $(c)$ to $(j)$ known in the literature?
If the reader is curious about the correctness of the results above and wants to verify them on Mathematica, the Mathematica command of $|E_r|$ is Abs[EulerE[r]]
Thanks,
Proof of (a): By using Euler's reflection formula $$\Gamma(m)\Gamma(1-m)=\pi \csc(m\pi),\quad m\notin\mathbb{Z}$$
and beta function
$$\operatorname{B}(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=\int_0^\infty \frac{x^{a-1}}{(1+x)^{a+b}}dx$$
with $a=m$ and $b=1-m$, we have
$$\pi\csc(m\pi)=\int_0^\infty\frac{x^{m-1}}{1+x}dx$$
differentiate both sides $2n$ times with respect to $m$ then let $m$ approach $1/2$
$$\int_0^\infty\frac{\ln^{2n}(x)}{1+x^2}dx=\frac{\pi}{2^{2n+1}}\lim_{m\to \frac12}\frac{d^{2n}}{dm^{2n}}\csc(m\pi)$$
the proof completes on using
$$\lim_{m\to \frac12}\frac{d^{2n}}{dm^{2n}}\csc(m\pi)=|E_{2n}|\pi^{2n}$$
which is explained here.
Proof of (b): Divide both sides of the common dilogarithm identity
$$\text{Li}_2(-z)+\text{Li}_2(-1/z)=-\frac{\ln^2(z)}{2}-2\eta(2)$$
by $z$ then integrate repeatedly.
Proof of (c): Let $yx^2=t^2$,
$$\int_0^\infty\frac{\ln^{2n}(x)}{1+yx^2}dx=\frac{1}{\sqrt{y}}\int_0^\infty\left(\ln(\sqrt{y})-\ln(t)\right)^{2n}\frac{dt}{1+t^2}$$
$$=\frac{1}{\sqrt{y}}\int_0^\infty\left(\sum_{k=0}^{2n} \binom{2n}{k}(-\ln(t))^{2n-k}\ln^k(\sqrt{y})\right)\frac{dt}{1+t^2}$$
$$=\frac1{\sqrt{y}}\sum_{k=0}^{2n}\binom{2n}{k}\ln^k(\sqrt{y})\int_0^\infty\frac{(-\ln(t))^{2n-k}}{1+t^2}dt$$
Since we care for only the even powers of $\ln(t)$ as the odd powers make the integral zero, we have
$$\sum_{k=0}^{2n}f(k)=\sum_{k=0}^n f(2k)$$
and so
$$\int_0^\infty\frac{\ln^{2n}(x)}{1+yx^2}dx=\frac1{\sqrt{y}}\sum_{k=0}^{n}\binom{2n}{2k}\ln^{2k}(\sqrt{y})\int_0^\infty\frac{\ln^{2n-2k}(t)}{1+t^2}dt$$
The proof completes on using the result in $(a)$.
Proof of (d): We follow the same steps in proof $(c)$:
$$\int_0^\infty\frac{\ln^{2n-1}(x)}{1+yx^2}dx=\frac{-1}{\sqrt{y}}\int_0^\infty\left(\ln(\sqrt{y})-\ln(t)\right)^{2n-1}\frac{dt}{1+t^2}$$
$$=\frac{-1}{\sqrt{y}}\int_0^\infty\left(\sum_{k=0}^{2n-1} \binom{2n-1}{k}(-\ln(t))^{2n-k-1}\ln^k(\sqrt{y})\right)\frac{dt}{1+t^2}$$
$$=\frac{-1}{\sqrt{y}}\sum_{k=0}^{2n-1}\binom{2n-1}{k}\ln^k(\sqrt{y})\int_0^\infty\frac{(-\ln(t))^{2n-k-1}}{1+t^2}dt$$
Since we care for only the even powers of $\ln(t)$, we have
$$\sum_{k=0}^{2n-1}f(k)=\sum_{k=0}^{n-1} f(2k+1)$$
and so
$$\int_0^\infty\frac{\ln^{2n-1}(x)}{1+yx^2}dx=\frac{-1}{\sqrt{y}}\sum_{k=0}^{n-1}\binom{2n-1}{2k+1}\ln^{2k+1}(\sqrt{y})\int_0^\infty\frac{\ln^{2n-2k-2}(t)}{1+t^2}dt$$
The proof completes on using the result in $(c)$.
Proof of (e): Using the integral representation of the polylogarithm function:
$$\text{Li}_{a}(z)=\frac{(-1)^{a-1}}{(a-1)!}\int_0^1\frac{z\ln^{a-1}(t)}{1-zt}dt$$
We have $$\int_0^\infty\frac{\ln^{2n}(x)\text{Li}_{2a+1}(-x^2)}{1+x^2}dx=\int_0^\infty\frac{\ln^{2n}(x)}{1+x^2}\left(\frac{1}{(2a)!}\int_0^1\frac{-x^2\ln^{2a}(y)}{1+yx^2}dy\right)dx$$
$$=\frac{1}{(2a)!}\int_0^1\ln^{2a}(y)\left(\int_0^\infty\frac{-x^2\ln^{2n}(x)}{(1+x^2)(1+yx^2)}dx\right)dy$$
$$=\frac{1}{(2a)!}\int_0^1\frac{\ln^{2a}(y)}{1-y}\left(\int_0^\infty\frac{\ln^{2n}(x)}{1+x^2}dx-\int_0^\infty\frac{\ln^{2n}(x)}{1+yx^2}dx\right)dy$$
use $(a)$ and $(c)$ for the inner integrals
$$=\frac{\left(\frac{\pi}{2}\right)^{2n+1}}{(2a)!}\left(|E_{2n}|\int_0^1\frac{\ln^{2a}(y)}{1-y}dy-\sum_{k=0}^n\binom{2n}{2k}|E_{2n-2k}|\pi^{-2k}\int_0^1\frac{\ln^{2k+2a}(y)}{\sqrt{y}(1-y)}dy\right)$$
The proof completes on using:
$$\int_0^1\frac{\ln^a(x)}{1-x}dx=(-1)^aa!\zeta(a+1)$$
$$\int_0^1\frac{\ln^a(x)}{\sqrt{x}(1-x)}dx=(-1)^aa!(2^{a+1}-1)\zeta(a+1)$$
Proof of (f): Following the same steps in proof $(e)$:
$$\int_0^\infty\frac{\ln^{2n-1}(x)\text{Li}_{2a}(-x^2)}{1+x^2}dx=\int_0^\infty\frac{\ln^{2n-1}(x)}{1+x^2}\left(\frac{1}{(2a-1)!}\int_0^1\frac{x^2\ln^{2a-1}(y)}{1+yx^2}dy\right)dx$$
$$=\frac{1}{(2a-1)!}\int_0^1\ln^{2a-1}(y)\left(\int_0^\infty\frac{x^2\ln^{2n-1}(x)}{(1+x^2)(1+yx^2)}dx\right)dy$$
$$=\frac{1}{(2a-1)!}\int_0^1\frac{\ln^{2a-1}(y)}{1-y}\left(\int_0^\infty\frac{\ln^{2n-1}(x)}{1+yx^2}dx-\int_0^\infty\frac{\ln^{2n-1}(x)}{1+x^2}dx\right)dy$$
substitute $(d)$ and notice that the second inner integral is zero
$$=\frac{-\left(\frac{\pi}{2}\right)^{2n-1}}{2(2a-1)!}\sum_{k=0}^{n-1}\binom{2n-1}{2k+1}|E_{2n-2k-2}|\pi^{-2k}\int_0^1\frac{\ln^{2k+2a}(y)}{\sqrt{y}(1-y)}dy$$
The proof completes on using
$$\int_0^1\frac{\ln^a(x)}{\sqrt{x}(1-x)}dx=(-1)^aa!(2^{a+1}-1)\zeta(a+1)$$
Proof of (g):
$$\int_0^1\frac{\ln^{2n}(x)\text{Li}_{2a+1}(-x^2)}{1+x^2}dx=\int_0^\infty\frac{\ln^{2n}(x)\text{Li}_{2a+1}(-x^2)}{1+x^2}dx-\underbrace{\int_1^\infty\frac{\ln^{2n}(x)\text{Li}_{2a+1}(-x^2)}{1+x^2}dx}_{x\to1/x}$$
$$=\int_0^\infty\frac{\ln^{2n}(x)\text{Li}_{2a+1}(-x^2)}{1+x^2}dx-\int_0^1\frac{\ln^{2n}(x)\text{Li}_{2a+1}(-1/x^2)}{1+x^2}dx$$
add the integral to both sides
$$2\int_0^1\frac{\ln^{2n}(x)\text{Li}_{2a+1}(-x^2)}{1+x^2}dx=\int_0^\infty\frac{\ln^{2n}(x)\text{Li}_{2a+1}(-x^2)}{1+x^2}dx$$ $$+\int_0^1\frac{\ln^{2n}(x)[\text{Li}_{2a+1}(-x^2)-\text{Li}_{2a+1}(-1/x^2)]}{1+x^2}dx$$
the first integral is given in $(e)$. For the second integral, replace $a$ by $2a+1$ in $(b)$ then use $\sum_{k=m}^na_k=\sum_{k=m}^{n}a_{n-k+m}$
$$\text{Li}_{2a+1}(-z)-\text{Li}_{2a+1}(-1/z)=-2\sum_{k=0}^a \frac{\eta(2a-2k)}{(2k+1)!}\ln^{2k+1}(z)$$
and so
$$\int_0^1\frac{\ln^{2n}(x)[\text{Li}_{2a+1}(-x^2)-\text{Li}_{2a+1}(-1/x^2)]}{1+x^2}dx=-4\sum_{k=0}^a\frac{\eta(2a-2k)}{(2k+1)!}4^k\int_0^1\frac{\ln^{2k+2n+1}(x)}{1+x^2}dx$$
and the proof completes on using
$$\int_0^1\frac{\ln^a(x)}{1+x^2}dx=(-1)^a a! \beta(a+1)$$
Proof of (h): We follow exactly the same steps in proof $(g)$ but here we use
$$\text{Li}_{2a}(-z)+\text{Li}_{2a}(-1/z)=-2\sum_{k=0}^a \frac{\eta(2a-2k)}{(2k)!}\ln^{2k}(z)$$
Proof of (i) and (j): Using the generating function
$$\sum_{k=1}^\infty H_k^{(a)} x^k=\frac{\text{Li}_a(x)}{1-x}$$
and the fact that
$$\int_0^1 x^{k}\ln^n(x)dx=\frac{(-1)^nn!}{(k+1)^{n+1}}$$
we have:
$$\sum_{k=1}^\infty\frac{(-1)^k H^{(2a+1)}_k}{(2k+1)^{2n+1}}=\frac1{(2n)!}\int_0^1\frac{\ln^{2n}(x)\text{Li}_{2a+1}(-x^2)}{1+x^2}dx$$
$$\sum_{k=1}^\infty\frac{(-1)^k H^{(2a)}_k}{(2k+1)^{2n}}=\frac{-1}{(2n-1)!}\int_0^1\frac{\ln^{2n-1}(x)\text{Li}_{2a}(-x^2)}{1+x^2}dx$$
These two integrals are given in $(g)$ and $(h)$.
Edit 3/26/2024
More generalized polylogarithmic integrals can be found in this preprint.
