# About the integral $\int_{0}^{+\infty}\frac{\text{arcsinh}(x)\,\text{arcsinh}(\lambda x)}{x^2}\,dx$

Context: I am working on a project involving integral means, algebraic inequalities and hypergeometric functions. Today I was studying the integral over the region $A<b<a<A+1$ of $$\int_{0}^{+\infty}\frac{dt}{\sqrt{(t^2+a^2)(t^2+b^2)}}$$ which is well-known to be related with the complete elliptic integral of the first kind and the AGM mean. My train of thoughts led me to the parametric integral $$I(\lambda) = \int_{0}^{+\infty}\frac{\text{arcsinh}(x)\,\text{arcsinh}(\lambda x)}{x^2}\,dx,\qquad \lambda>0$$ which my version of Mathematica returns as a Meijer G function, and by other ways I know to be related with a series involving squared central binomial coefficients and values of the incomplete Beta function. $I(1)=\frac{\pi^2}{2}$ is straightforward to prove.

Question: Is it possible to find a closed form for $I(\lambda)$ in terms of "usual" functions, or at least a manageable representation as a fast-convergent series?

My first temptation was to apply Feyman's trick, but $\int_{0}^{+\infty}\frac{\text{arcsinh}(x)}{x\sqrt{1+\lambda^2 x^2}}\,dx$ does not seem really easier to tackle or to integrate. I hope that I am wrong, of course. There also is a similar integral which has a simple closed form: $$\int_{0}^{+\infty}\frac{\log(1+x^2)\log(\lambda^2+x^2)}{x^2}\,dx = 2\pi\left(1+\tfrac{1}{\lambda}\right)\log(\lambda+1).$$

• $I(\lambda)$ is closely related to $< f, g >_{\mathbb{L}_2}$, where $f = \frac{\text{arcsinh}(x)\,\text{arcsinh}(\lambda x)}{x}$ and $g = \frac{1}{x}$. What happens if you consider $< \hat{f}, \hat{g}>$, meaning by hat the Fourier transform? This will preserve the scalar product and hopefully might be easier to calculate. Any other isometry could be an attempt. – Ivan Di Liberti Aug 25 '17 at 0:04
• @IvanDiLiberti: the Fourier transform of $\frac{1}{x}$ is singular and to compute the Fourier transform of $f$ does not seem to be easier than computing $I(\lambda)$, however. On the other hand, we may get luckier with the Laplace transform, thanks for the suggestion :D – Jack D'Aurizio Aug 25 '17 at 0:07
• The Laplace transform of $\text{arcsinh}(x)$ is given by a combination of a Bessel Y and a Struve H functions, if someone is willing to take this road. – Jack D'Aurizio Aug 25 '17 at 0:09
• I was passing by, a comment was duty. – Ivan Di Liberti Aug 25 '17 at 0:10
• @JackD'Aurizio You may be interested in this paper: arxiv.org/abs/math/0601082 . It's a fantastic bit expositional mathematics showing the application of polylog functions in the solution of a certain class of inverse-trig integrals, and their relations to Mahler measures. – David H Aug 28 '17 at 15:51

From the generalized Parseval relation for the Mellin transform, we can use the result $$\int_0^\infty g(\lambda x)g(x)x^{-2}\,dx=\frac{1}{2i\pi}\int_{\delta-i\infty}^{\delta+i\infty}\tilde{g}(s)\tilde{g}(-1-s)\lambda^{-s}\,ds$$ Here, from Ederlyi table (6.6.13) p.323, for $g(x)=\text{arcsinh} (x)$, one has $$\tilde{g}(s)=-\frac{1}{2s}B\left( \frac{s+1}{2},\frac{-s}{2} \right)$$ for $-1<s<0$. Then $$I(\lambda)=\frac{-1}{8i\pi^2}\int_{\delta-i\infty}^{\delta+i\infty}\Gamma^2\left( \frac{1+s}{2} \right)\Gamma^2\left( -\frac{s}{2} \right)\frac{\lambda^{-s}}{s(s+1)}\,ds$$ with $-1<\delta<0$. Poles are at $s=-1,-3,-5...$ and $0,2,4...$. Poles at $s=-1,0$ are of order 3 the other are double. For $s\to\infty$, the function to be integrated is $\sim \left|\lambda\right|^{-s}s^{-3}\csc^{-2}(\pi s/2)$. Then, for $\left|\lambda\right|<1$, we close the contour with a large half-circle on the left of the vertical line $\Re(s)=\delta$. With the help of Maple to compute the residues, one gets \begin{align} I(\lambda)=\frac{1}{2}&\left( \left( \ln \frac{\lambda}{4}-1 \right)^2 +1+\frac{\pi^2}{3}\right)\lambda -\frac{1}{2\pi}\sum_{n=2}^\infty\left( \frac{\Gamma\left( n-\frac{1}{2} \right)}{\Gamma(n)} \right)^2.\\ &.\left[-\ln(\lambda)+ \psi(n+1)-\psi(n+\frac{1}{2})+\frac{4n^2-n-2}{2n(n-1)(2n-1)} \right]\frac{\lambda^{2n-1}}{(n-1)(2n-1)} \end{align} The series converges for $\left|\lambda\right| < 1$.

For $\left|\lambda\right| >1$, we may use the relation $$I(\lambda)=\lambda I\left(\frac{1}{\lambda}\right)$$ which is obvious from the integral expression. This functional relation may be retrieved by closing the contour by a half-circle on the right of the line.

• how do you close the contour of integration? – tired Aug 28 '17 at 22:46
• I should have given this. I added information on the contour. With a large half circle on the left of the vertical line, we describe the region $\lambda <1$. Closing to the right gives $\lambda>1$. – Paul Enta Aug 29 '17 at 6:07
• thanks! nice work +1 – tired Aug 30 '17 at 0:06

This is not going to be the full answer to this question because I believe that this integral cannot be represented in terms of elementary functions plus poly-logarithms. I will simply extract the "simple" part out of this integral and leave the undigested part as it is in hope that some genius might tackle it in the future. We start by computing the anti-derivative of the integrand. Firstly we note the following result:

\begin{eqnarray} (1) \int \frac{\log(x)}{\sqrt{1+\lambda^2 x^2}}dx=\frac{1}{2 \lambda} \left[\mbox{arcsinh}(\lambda x) \left( \mbox{arcsinh}(\lambda x) - 2 \log(2 \lambda)\right) + Li_2(e^{-2\mbox{arcsinh}(\lambda x)})\right] \end{eqnarray}

We have: \begin{eqnarray} &&\int \frac{\mbox{arcsinh}(x) \mbox{arcsinh}(\lambda x)}{x} dx=\\ && -\frac{1}{x} \mbox{arcsinh}(x) \mbox{arcsinh}(\lambda x) +\\ && (\log(x)-\log(1+\sqrt{1+x^2})) \mbox{arcsinh}(\lambda x) + (\log(x)-\log(1+\sqrt{1+\lambda^2 x^2})) \mbox{arcsinh}( x) \lambda +\\ &&-\lambda \int\frac{\log(x)-\log(1+\sqrt{1+x^2})}{\sqrt{1+\lambda^2 x^2}}dx -\lambda \int\frac{\log(x)-\log(1+\sqrt{1+\lambda^2 x^2})}{\sqrt{1+ x^2}}dx=\\ &&\frac{1}{2} \left(\right.\\ &&\left.-\lambda \text{Li}_2\left(e^{-2 \mbox{arcsinh}(x)}\right)-\text{Li}_2\left(e^{-2 \mbox{arcsinh}(\lambda x)}\right)\right.\\ &&\left.-2 \log \left(\sqrt{x^2+1}+1\right) \mbox{arcsinh}(\lambda x)-2 \lambda \mbox{arcsinh}(x) \log \left(\sqrt{\lambda^2 x^2+1}+1\right)-\lambda \mbox{arcsinh}(x)^2\right.\\ &&\left.-\frac{2 \mbox{arcsinh}(x) \mbox{arcsinh}(\lambda x)}{x}-\mbox{arcsinh}(\lambda x)^2\right.\\ &&\left.+2 \lambda \log (x) \mbox{arcsinh}(x)+\lambda \log (4) \mbox{arcsinh}(x)+2 \log (2 \lambda) \mbox{arcsinh}(\lambda x)+2 \log (x) \mbox{arcsinh}(\lambda x)\right.\\ &&\left.\right)+\\ &&\lambda \underbrace{\int \frac{\log(1+\sqrt{1+x^2})}{\sqrt{1+\lambda^2 x^2}} dx}_{{\mathcal I}_1(x)}+\lambda \underbrace{\int \frac{\log(1+\sqrt{1+\lambda^2 x^2})}{\sqrt{1+ x^2}} dx}_{{\mathcal I}_2(x)} \end{eqnarray} In the first line we integrated by parts twice and in the second line we used the identity(1). Now, we deal with the remaining integrals by expanding them in a Taylor series about $\lambda=1$. We have:

\begin{eqnarray} &&{\mathcal I}_1(x) = -\frac{1}{2} [\log(2 z)]^2 - 2 Li_2(-z) +\\ && (\lambda-1) \cdot \left( 2 \arctan(z)+2 Li_2(-z)+\frac{1}{2} \log(z)^2+\right.\\ &&\left.\frac{\left(2 \log (2)-z^2 (4-2 \log (2))\right) \log (z)+4 \left(z^2-1\right) \log (z+1)+4 \log (2)}{2 \left(z^2+1\right)}\right)+O((\lambda-1)^2)\\ &&{\mathcal I}_2(x) = -\frac{1}{2} [\log(2 z)]^2 - 2 Li_2(-z) +\\ && (\lambda-1) \cdot \left( -2 \arctan(z)+\log(z)\right) + O((\lambda-1)^2) \end{eqnarray}

where $z:= \exp(\mbox{arcsinh}(x))$.