For naturals $n\ge m$, define $$I(n,m):=\int_0^\frac12\dfrac{\text{arcsinh}^nx}{x^m}dx$$

with $\text{arcsinh}\ x=\ln(x+\sqrt{1+x^2} )$, so $\text{arcsinh} \frac12=\ln \frac{\sqrt{5}+1}2 $.
Is it possible to find closed form expressions of $I(n,m)$? I mean closed form in a broad sense, i.e. involving any other "known" constants.

Motivation: it is known that $$I(2,1)=\int_0^\frac12\dfrac{\text{arcsinh}^2x}xdx=\dfrac{\zeta(3)}{10},$$

further $$I(1,0)= \int_0^\frac12 \text{arcsinh}\ x\ dx=\frac12\left(2-\sqrt{5}+\ln \frac{\sqrt{5}+1}2 \right)$$ $$I(1,1)= \int_0^\frac12\dfrac{\text{arcsinh}\ x}xdx=\frac{\pi^2}{20}$$ and $$I(2,2)=\int_0^\frac12\dfrac{\text{arcsinh}^2x}{x^2}dx=\frac{\pi^2}6-5\ln^2\frac{\sqrt{5}+1}2 $$ so there might be some hope that at least some others of the $I(n,m)$ have closed forms involving values of $\zeta(k)$, ideally odd zeta values.

  • $\begingroup$ One obtains similar results, by replacing $\arcsinh$ with $\arcsin$ and integrating from 0 to 1: $J(1,1)=\frac{\pi}{2}\ln 2, J(2,1)=-\frac{7}{8}\zeta(3)+\frac{\pi^2}{4}\ln 2$, $J(2,2)=4C-\frac{\pi^2}{4}$... It seems that $J(2n+1)$ contains $\zeta(2n+1)$. Moreover simple series exist for the smaller values of $m$ and $J(n,0)$ is "polynomial" in $\pi$. $\endgroup$ – Paul Enta Feb 15 '16 at 22:06
  • $\begingroup$ Mathematica gets expressions involving polylogs in all cases I tried (but not general closed form) $\endgroup$ – Igor Rivin Feb 15 '16 at 22:30

The following results are quoted in http://www.hindawi.com/journals/ijmms/2007/019381/abs/ (Integer Powers of Arcsin, by J.M. Borwein and M. Chamberland): $$\large I(4,1)=-\frac{3}{2}\mathrm{Li}_5(g^2)+3\mathrm{Li}_4(g^2)\ln{g}+\frac{3}{2}\zeta(5)-\frac{12}{5}\zeta(3)\ln^2{g}-\frac{4}{15}\pi^2\ln^3{g}+\frac{4}{5}\ln^5{g},$$ where $g=(\sqrt{5}-1)/2$ is the golden ratio, and $$\large \frac{2^n}{n!}I(n,1)=\zeta(n+1)-\frac{n(-\ln{g^2})^{n+1}}{2(n+1)!}-\sum\limits_{j=2}^{n+1}\frac{(-\ln{g^2})^{n+1-j}}{(n+1-j)!}\mathrm{Li}_j(g^2).$$

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Letting $y=\text{arcsinh}\, x$ and integrating by parts, we have $$I(n,m)=-\frac{2^{m-1}a^n}{m-1}+\frac n{m-1}\,[J(n-1,m-1;a)-J(n-1,m-1;0)] $$ if $n\ge m\ge2$, where $a:=\text{arcsinh}\,\frac12$ and $$J(p,q;y):=\int\frac{y^p}{\sinh^qy}\,dy.$$ Formula in Prudnikov--Brychkov--Marichev (PBM, Vol. 1, ISBN 5-9221-0323-7) tells us that
$$J(p,q;y)= -\frac{py^{p-1}}{(q-1)(q-2)\sinh^{q-2}y} -\frac{y^p\cosh y}{(q-1)\sinh^{q-1}y}$$ $$ +\frac{p(p-1)}{(q-1)(q-2)}\,J(p-2,q-2;y) -\frac{q-2}{q-1}\,J(p,q-2;y). $$ Also, formulas and in PBM tell us that $$J(p,1;y)=\sum_{k=0}^\infty\frac{(2-2^{2k})B_{2k}}{(2k)!(p+2k)}\,y^{p+2k}\quad (|y|<\pi,\ p>0)$$ and $$J(p,2;y)=-y^p\coth y+p\sum_{k=0}^\infty\frac{2^{2k}B_{2k}}{(2k)!(p+2k-1)}\,y^{p+2k-1}\quad (|y|<\pi,\ p>1).$$ These formulas in PBM should be easy to obtain/check.

Since $|a|=a=\text{arcsinh}\,\frac12<1/2<\pi$, the above formulas provide a recursion to compute the values of $I(n,m)$ in terms of the Bernoulli numbers $B_{2k}$ or, alternatively, in terms of $J(p,1;a)-J(p,1;0)$ and $J(p,2;a)-J(p,2;0)$, that is, in terms of $I(n,2)$ and $I(n,3)$.

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  • $\begingroup$ Do you mean J(p,q;y) to be the definite integral from 0 to y? $\endgroup$ – Wolfgang Feb 24 '16 at 9:19
  • $\begingroup$ Wolfgang: $J(p,q,y)$ is, as written, the indefinite integral, that is, the set of all antiderivatives of the corresponding integrand. The recursion for $J(p,q,y)$ is then understood using the Minkowski operations over sets, so that $c+sA+tB=\{c+sa+tb\colon a\in A,b\in B\}$ for any real $c,s,t$ and any subsets $A,B$ of $\mathbb R$. $\endgroup$ – Iosif Pinelis Feb 24 '16 at 14:34
  • $\begingroup$ @Pinelis: E.g. the series of $J(p,1;y)$ and $J(p,2;y)$ can be represented by "Multiple Gammafunction" (closed form). $\endgroup$ – user90369 May 6 '16 at 8:44


$$ I(3,3)=-2\, \left( {\it arcsinh} \left( 1/2 \right) \right) ^{3}-3/2\, \left( {\it arcsinh} \left( 1/2 \right) \right) ^{2}\sqrt {5}+{ \frac {3}{20}}\,{\pi }^{2}$$

This holds to at least 500 decimal digits.

Observation: If negative $m$ is allowed, sage finds closed form, e.g.:

sage: i1=integral(asinh(x)^10/x^(-10),x,0,1/2);i1
234457/609840000000*(390625*sqrt(5))*arcsinh(1/2)^9 + 736376357621/4067709030000000*(78125*sqrt(5))*arcsinh(1/2)^7 + 9180625625588424869/46512218903535000000*(3125*sqrt(5))*arcsinh(1/2)^5 + 1/22528*arcsinh(1/2)^10 - 874194067/601058304*arcsinh(1/2)^8 + 110751609620625434006951381/5584361654050945053750000*(625*sqrt(5))*arcsinh(1/2)^3 - 12917439496549363/128865022070400*arcsinh(1/2)^6 - 159325079188097380354387/51572748320239608000*arcsinh(1/2)^4 + 1330454347355585383829820394548869/446980683332718718519730625000*(25*sqrt(5))*arcsinh(1/2) - 1919167341505471379018435677363/51599501683430732296650000*arcsinh(1/2)^2 - 23050668024232648435666069406694778387/309757613549574071934173323125000
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  • $\begingroup$ For negative $m$, integration by parts yields a recursion leaving integrals of the form $ \int\frac{x^k \ dx}{\sqrt{1+x^2}} $, so it is obvious that even the indefinite integral has a closed form then. $\endgroup$ – Wolfgang Feb 16 '16 at 9:11
  • $\begingroup$ @Wolfgang OK. The conjecture is very close to the form for negative $m$, except for $\pi$. $\endgroup$ – joro Feb 16 '16 at 9:34

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