# Are these generalizations known in the literature?

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 $$(h)$$.

• (+1) Great pack of results to have in hand and refer to! :-) Mar 6, 2022 at 7:00
• Very impressive. Mar 7, 2022 at 4:33
• (+1) Great work. Nice proofs, and nicely laid out, as you always do. I'll put a printout into my copy of your book. Mar 8, 2022 at 9:48
• Oh yes, now these generalizations are known in the (online) literature !! Mar 10, 2022 at 12:59
• It might be interesting to know that these results have been submitted virtually literally by a group of authors to a scientific journal 25 days after the publication here and have been published without mentioning the source. I wrote an e-mail to the responsible author pointing out this violation of the scientific code of conduct. He replied finally, quote, "I have e-mailed the editor in chief of the journal asking him to retract the paper." Sep 14, 2022 at 16:50