# How to prove that $\int _0^\infty\frac{\text{arcsinh}^nx}{x^m}dx$ is a rational combination of zeta values?

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 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: The integrals $$J(n,m):= \int _0^1\dfrac{\text{arctanh}^nx}{x^m}dx= \int _0^1\log^{n/2}\left(\dfrac{1+x }{1-x}\right)\dfrac{dx}{x^m}$$ and $$K(n,m):= \int _1^\infty\dfrac{\text{arcoth}^nx}{x^m}dx =\int _1^\infty\log^{n/2}\left(\dfrac{x+1}{x-1}\right)\dfrac{dx}{x^m}$$ have very similar situations with the exact same zeta values involved as for $I(n,m)$:
We have $$J(n,m)=\frac{n!}{2^{n-m}m!} \cdot Z^{n}p_ {m}(\frac1{Z} ),\qquad K(n,m)=\frac{n!}{2^{n-m}m!} \cdot\bar Z^{n}p_ {m}(\frac1{\bar Z} )$$ with even/odd polynomials $p_m$, the same for both integrals, of degree $m−2$. Again, powers $Z^r$ are to be replaced by $\zeta(r)$, but powers $\bar Z^r$ are to be replaced this time by $\bar\zeta(r):=\dfrac{2^{r-1}-1}{2^{r-1}}\zeta(r)$.

Here the polynomials $p_m\equiv p_m(z)$ are (essentially) not reducible, and instead of a recursion for $J(n,m)$, there is at least one for the polynomials: $$p_1=0,\ p_2=1, \ \frac{p_m}m=z\frac {p_{m-1}}{m-1} +\frac {m-3}4p_{m-2}.$$ The first of them are
$p_3=\frac32z$
$p_4=2z^2+1$
$p_5=\frac52( z^3+2z)$
$p_6=\frac32(2z^4+10z^2+3)$
$p_7=\frac74(2z^5+20z^3+23z)$
$p_8=4z^6 +70z^4+196z^2+45$
$p_9=\frac92(z^7+28z^5+154z^3+132z)$

Note that $p_m(1)=\dfrac{m!}{2^{m-1}}$ for $m\ge2$.

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.

• Thank you! For the record: the integral $J(n,m):= \int _0^1\dfrac{\text{arctanh}^nx}{x^m}dx$ has a very similar situation with the exact same zeta values involved as for $I(n,m)$: We have $J(n,m)=\frac{n!}{m!2^{n-m}} \cdot Z^{n}p_m(\frac1{Z} )$, with even/odd polynomials $p_n$ of degree $n-2$. But here the $p_n$ are not reducible, and instead of a recursion for $J(n,m)$, there is at least one for the polynomials: $p_1=0,p_2=1, \frac{p_n}n=Z\frac {p_{n-1}}{n-1} +\frac {n-3}4p_{n-2}$. E.g. $p_6=\frac32(2Z^4+10Z^2+3),p_7=\frac74(2Z^5+20Z^3+23Z).$ – Wolfgang Feb 26 '16 at 20:34

Letting $y=\text{arcsinh}\, x$ and integrating by parts, we have $$I(n,m)=\frac n{m-1}\,J(n-1,m-1)$$ if $n\ge m\ge2$, where $$J(p,q):=\int_0^\infty\frac{y^p}{\sinh^qy}\,dy.$$ Formula 1.4.24.1 in Prudnikov--Brychkov--Marichev (PBM, Vol. 1, ISBN 5-9221-0323-7) implies
$$J(p,q)= \frac{p(p-1)}{(q-1)(q-2)}\,J(p-2,q-2) -\frac{q-2}{q-1}\,J(p,q-2)$$ if $p\ge q>2$.
That formula in PBM (which is reproduced in my answer at Is there a closed form of $\int_0^\frac12\dfrac{\text{arcsinh}^nx}{x^m}dx$?) should be easy to obtain/check. Thus, we have a recursion to reduce the value of $I(n,m)$ to those of $J(p,1)$ and $J(p,2)$, that is, to those of $I(n,2)$ and $I(n,3)$.

• Actually I proved the OP's conjecture by a direct calculation. See my response. – GH from MO Feb 23 '16 at 23:38
• GH from MO: I see. I am wondering if this rather simple recursion makes it possible to check the conjecture by induction. I'll try to find time to do that. – Iosif Pinelis Feb 24 '16 at 1:08
• Yes, the recursion becomes $(q-1)(q-2)I(p,q)=(p-1)pI(p-2,q-2)-(q-3)^2I(p,q-2)$. – Wolfgang Feb 24 '16 at 9:50