For $n\ge m\ge 2$, define $$I(n,m):= \int \_0^\infty\dfrac{\text{arcsinh}^nx}{x^m}dx$$ Computer algebra systems say that the indefinite integral can be expressed in terms of polylog functions (of rapidly increasing complexity), but I see no way of doing the limit $\to\infty$. On the other hand, I have found numerically that $I(n,m)$ can (conjecturally always) be written as a rational combination of zeta values, moreover **this happens in a rather beautiful way**.


Here are the values for small $m$
and then what seems to be the general formula.  

Defining $\bar\zeta(n):=(2^n-1) \zeta(n)$, it seems like


$ I(n,2)=\dfrac{n!}{2^{n-1}} \bar\zeta(n)$


$ I(n,4)=\dfrac{n!}{2^{n-1}} \cdot\dfrac1 {6}[ 4\bar\zeta(n-2)-\bar\zeta(n)]$


$ I(n,6)=\dfrac{n!}{2^{n-1}} \cdot\dfrac1{120}[ 16\bar\zeta(n-4)-40\bar\zeta(n-2)+9\bar\zeta(n)]$


$ I(n,8)=\dfrac{n!}{2^{n-1}} \cdot\dfrac1{5040}[ 64\bar\zeta(n-6)-560\bar\zeta(n-4) +1036\bar\zeta(n-2)-225\bar\zeta(n)]$


*(you may want to stop here for a moment before reading on and look if the coefficients tell you something already)*


and


$ I(n,3)=\dfrac{n!}{2^{n-1}} \zeta(n-1)$


$ I(n,5)=\dfrac{n!}{2^{n-1}} \cdot\dfrac13[ \zeta(n-3)-\zeta(n-1)]$


$ I(n,7)=\dfrac{n!}{2^{n-1}} \cdot\dfrac2{45}[\zeta(n-5)- 5 \zeta(n-3)+4\zeta(n-1)]$


Well, in [umbral notation](https://en.wikipedia.org/wiki/Umbral_calculus) introducing a formal variable $\bar Z$, for even $m=2k$ $$ I(n,2k)=\frac{n!}{2^{n-1}} \cdot\frac1{(2k-1)!}\bar Z^{n}\prod\_{j=-k+1}^{k-2} \left(\frac2{\bar Z}-(2j+1)\right)$$ where each power $\bar Z^r$ is to be replaced by $\bar\zeta (r)$. E.g. for $m=6$, the "zeta polynomial" is related to $16z^4-40z^2+9=(2z-3)(2z-1)(2z+1)(2z+3)$.


Likewise for odd $m=2k+3$ (*sic* to make notation more elegant), $$ I(n,2k+3)=\frac{n!}{2^{n-1}} \cdot\frac{2^{2k}}{(2k+1)!}Z^{n}\prod\_{j=-k}^{k}\left (\frac1{Z}-j\right)$$ where each power $Z^r$ is to be replaced by $\zeta (r)$. E.g. for $m=7$, the "zeta polynomial" is related to $z^5-5z^3+4z=(z-2)(z-1)z(z+1)(z+2)$.


We can rewrite the expression for even $m$ as follows to look amazingly similar to the one for odd $m$:


$$ I(n,2k+2)=\frac{n!}{2^{n-1}} \cdot\frac{2^{2k}}{(2k+1)!}\bar Z^{n}\prod\_{j=-k+1/2}^{k-1/2} \left(\frac1{\bar Z}-j\right)$$ where in the product $j$ runs over the half-integers.



> 
> So far, this is all speculation, but once these patterns found, it seems obvious that there must be some deeper connection (which might even yield an elegant proof). Any insights? Has anybody encountered similar families of integrals?
> 
> 
> 


**EDIT for the record:**
A similar pattern occurs for the integral
$$J(n,m):=\int \_0^\infty\dfrac{x^n}{\cosh^{m}x}dx,$$ which features for even $m$ the [Dirichlet eta function](https://en.wikipedia.org/wiki/Dirichlet_eta_function) $\eta(s)= \sum\limits\_{n=1}^{\infty}\dfrac {(-1)^{n-1} }{ n^s}= \left(1-2^{1-s}\right) \zeta(s)$,
and for odd $m$ the [Dirichlet beta function](https://en.wikipedia.org/wiki/Dirichlet_beta_function) $\beta(s) = \sum\limits\_{n=0}^\infty \dfrac{(-1)^n} {(2n+1)^s}$.


Using the same umbral notation as above for $\eta$ and $\beta$ respectively, we have for even $m=2k$


$$\bbox[10px,#cFF]{ \int \_0^\infty\dfrac{x^n}{\cosh^{2k}x}dx=\frac{2^{2k-n}~n!}{(2k-1)!} \eta^{n} \prod\_{j=1}^{k-1}\left ( j^2-\frac { 1}{ \eta^2}\right) } $$ and for odd $m=2k+1$


$$\bbox[10px,#cFF]{\int \_0^\infty\dfrac{x^n}{\cosh^{2k+1}x}dx=2\frac{n!}{(2k)!}\beta^n \prod\_{j=1}^{k} \left ( (2j-1)^2-\frac { 1}{ \beta^2}\right).} $$


For example, if $m=8$, we get from $(x^2-1)(4x^2-1)(9x^2-1)=36x^6-49x^4+14x^2-1$ (note that these polynomials for even $m$ are essentially the same as the ones in the case of odd $m$ for the original integrals $I(n,m)$ above)
$$J(n,8)=\int \_0^\infty\dfrac{x^n}{\cosh^{8}x}dx=\frac{n!}{7!~2^{n-8}}\Bigl[36~\eta(n)-49~\eta(n-2)+14~\eta(n-4)-\eta(n-6)\Bigr]$$


and from $(x^2-1)(9x^2-1)(25x^2-1)=225x^6-259x^4+35x^2-1$, we get for $m=7$
$$J(n,7)=\int \_0^\infty\dfrac{x^n}{\cosh^{7}x}dx=2\frac{n!}{6!} \Bigl[225~\beta(n)-259~\beta(n-2)+35~\beta(n-4)-\beta(n-6)\Bigr]$$



How to prove that $\int _0^\infty\frac{\text{arcsinh}^nx}{x^m}dx$ is a rational combination of zeta values?
Let me first establish the case $m=2$. The general case is established in the "Added" section below.


By making the substitution $x=\sinh t$ we get
\begin{align\*}
I(n,2)&=\int\_0^\infty t^n\frac{\cosh t}{\sinh^2 t}\,dt
=2\int\_0^\infty t^n\frac{e^t+e^{-t}}{(e^t-e^{-t})^2}dt\\
&=2\int\_0^\infty t^n\left(\sum\_{r=0}^\infty(2r+1)e^{-(2r+1)t}\right)dt\\
&=2\sum\_{r=0}^\infty(2r+1)\left(\int\_0^\infty t^n e^{-(2r+1)t}dt\right)\\
&=2\sum\_{r=0}(2r+1)^{-n}\left(\int\_0^\infty t^n e^{-t}dt\right)\\
&=2n!\sum\_{r=0}(2r+1)^{-n}=2n!(1-2^{-n})\zeta(n)=\frac{n!}{2^{n-1}} \bar\zeta(n).
\end{align\*}


**Added.** Here is the general case for $m=2k$ even. In the umbral notation, we need to replace $(\bar Z/2)^r$ by
$$ 2^{-r}\bar\zeta (r)=(1-2^{-r})\zeta(r)=\sum\_{s\text{ odd}}s^{-r}, $$
hence the conjectured formula really means
$$ I(n,2k)\overset{?}=\frac{2n!}{(2k-1)!}\sum\_{s\text{ odd}}s^{-n}\prod\_{j=-k+1}^{k-2}\bigl(s-(2j+1)\bigr).$$
The inside product is a polynomial of degree $2k-2$, and we see that it vanishes for $s\leq 2k-3$. Hence we make the substitution $s=2r+2k-1$, and we rewrite the conjectured formula as
$$ I(n,2k)\overset{?}=n!\sum\_{r=0}^\infty (2r+2k-1)^{-n} \frac{2^{2k-1}}{2k-1}\binom{r+2k-2}{2k-2}.$$
We prove this equation. From the definition, similarly as in the basic case $k=1$ above, we find
\begin{align\*}
I(n,2k)&=2^{2k-1}\int\_0^\infty t^n e^{(1-2k)t}(1+e^{-2t})(1-e^{-2t})^{-2k}dt\\
&=2^{2k-1}\int\_0^\infty t^n e^{(1-2k)t} \sum\_{r=0}^\infty\binom{-2k}{r}(-1)^r \left(e^{-2rt}+e^{-2(r+1)t}\right)dt\\
&=2^{2k-1}\int\_0^\infty t^n e^{(1-2k)t}\left(1+\sum\_{r=1}^\infty\left(\binom{-2k}{r}-\binom{-2k}{r-1}\right)(-1)^r e^{-2rt}\right)dt\\
&=2^{2k-1}\int\_0^\infty t^n e^{(1-2k)t}\left(1+\sum\_{r=1}^\infty\left(\binom{r+2k-1}{r}+\binom{r+2k-2}{r-1}\right)e^{-2rt}\right)dt\\
&=2^{2k-1}\int\_0^\infty t^n \sum\_{r=0}^\infty\left(\binom{r+2k-1}{2k-1}+\binom{r+2k-2}{2k-1}\right)e^{-(2r+2k-1)t}dt\\
&=2^{2k-1}\sum\_{r=0}^\infty \left(\binom{r+2k-1}{2k-1}+\binom{r+2k-2}{2k-1}\right) \int\_0^\infty t^ne^{-(2r+2k-1)t}dt\\
&=2^{2k-1}\sum\_{r=0}^\infty \left(\binom{r+2k-1}{2k-1}+\binom{r+2k-2}{2k-1}\right)\frac{n!}{(2r+2k-1)^{n+1}}\\
&=2^{2k-1}\sum\_{r=0}^\infty \frac{2r+2k-1}{2k-1}\binom{r+2k-2}{2k-2}\frac{n!}{(2r+2k-1)^{n+1}}\\
&=n!\sum\_{r=0}^\infty (2r+2k-1)^{-n}\frac{2^{2k-1}}{2k-1}\binom{r+2k-2}{2k-2}.
\end{align\*}
The proof is complete for the case of even $m$. 


The case of odd $m$ is similar, so I omit the details.


