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

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 \begin{equation} 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 \end{equation} 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.