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I was having fun working out integrals, and I came across this one:

$$\int \ln (x+\sqrt{x^2+1})\cdot\ln(x+\sqrt{x^2-1})\mathrm d x$$

I used integration by parts twice and ended up with this:

$$\int\left(\sqrt{\dfrac{x^2-1}{x^2+1}}+\sqrt{\dfrac{x^2+1}{x^2-1}}\right)\mathrm d x$$ Now I'm really stuck. Could you help me continue?

EDIT: Steps:
$\int \ln (x+\sqrt{x^2+1})\cdot\ln(x+\sqrt{x^2-1})\mathrm d x$
$=x\cdot\ln (x+\sqrt{x^2+1})\cdot\ln(x+\sqrt{x^2-1})-\int x\mathrm d \ln (x+\sqrt{x^2+1})\cdot\ln(x+\sqrt{x^2-1})$
$=x\cdot\ln (x+\sqrt{x^2+1})\cdot\ln(x+\sqrt{x^2-1})-\int x (\dfrac{\ln (x+\sqrt{x^2+1})}{\sqrt{x^2-1}}+\dfrac{\ln(x+\sqrt{x^2-1})}{\sqrt{x^2+1}})\mathrm d x$
$=x\cdot\ln (x+\sqrt{x^2+1})\cdot\ln(x+\sqrt{x^2-1})-\sqrt{x^2-1}\ln (x+\sqrt{x^2+1})-\sqrt{x^2+1}\ln(x+\sqrt{x^2-1})+\color{red}{\int\left(\sqrt{\dfrac{x^2-1}{x^2+1}}+\sqrt{\dfrac{x^2+1}{x^2-1}}\right)\mathrm d x}$
$=\cdots$

EDIT: This question is not so "meaningless". The function in question indeed exists, and according to Desmos, it is continuous on (1, +∞).

Scratch---Cat
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  • The integral over $]1,+\infty[$ does not converge. On the other hand, $I=\int\ln(x+\sqrt{x^{2}+1})\ln(x+\sqrt{x^{2}-1}){\rm d}x$ can be calculated in terms of elliptical integral $\int_{0}^{x}G(t,\sqrt{P(t)}){\rm d}t$. –  Jul 14 '21 at 10:11
  • I don't know why you would accept someone's programming facility for verifying the domain of something this straightforward. In particular, note that $\sqrt{x^2-1}$ is undefined for $-1 < x < 1,$ so without any further analysis we know the domain CAN'T be all of the interval $(0,\infty).$ – Dave L. Renfro Jul 14 '21 at 10:12
  • Can we simplify your second rendering of the integral to get$$\int\left(\dfrac{2x^2}{\sqrt{x^4-1}}\right)\mathrm d x\enspace ?$$ – YNK Jul 14 '21 at 10:15
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    Maybe the fact that $\mathrm{sinh}^{-1}x = \ln(x + \sqrt{x^{2} + 1})$ and $\mathrm{cosh}^{-1}x = \ln(x + \sqrt{x^{2} - 1})$ can help. – soupless Jul 14 '21 at 10:54
  • Could you show the steps leading to the last expression ? – Claude Leibovici Jul 14 '21 at 13:20
  • Very cool how this integral looks like it's going to lead to polylogarithms but quickly takes a detour to elliptic integrals instead. :) – David H Jul 18 '21 at 23:19

1 Answers1

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Define the function $\mathcal{I}:\left[1,\infty\right)\rightarrow\mathbb{R}_{\ge0}$ via the definite integral

$$\mathcal{I}{\left(z\right)}:=\int_{1}^{z}\mathrm{d}x\,\ln{\left(x+\sqrt{x^{2}+1}\right)}\ln{\left(x+\sqrt{x^{2}-1}\right)}.$$

Obviously, $\mathcal{I}{\left(z\right)}$ vanishes trivially at $z=1$.

Suppose $z>1$. As has been pointed out, this logarithmic integral can be reduced to one with a much simpler algebraic integrand through repeated integration by parts:

$$\begin{align} \mathcal{I}{\left(z\right)} &=\int_{1}^{z}\mathrm{d}x\,\ln{\left(x+\sqrt{x^{2}+1}\right)}\ln{\left(x+\sqrt{x^{2}-1}\right)}\\ &=z\ln{\left(z+\sqrt{z^{2}+1}\right)}\ln{\left(z+\sqrt{z^{2}-1}\right)}\\ &~~~~~-\int_{1}^{z}\mathrm{d}x\,x\frac{d}{dx}\left[\ln{\left(x+\sqrt{x^{2}+1}\right)}\ln{\left(x+\sqrt{x^{2}-1}\right)}\right];~~~\small{I.B.P.}\\ &=z\ln{\left(z+\sqrt{z^{2}+1}\right)}\ln{\left(z+\sqrt{z^{2}-1}\right)}\\ &~~~~~-\int_{1}^{z}\mathrm{d}x\,x\left[\frac{\ln{\left(x+\sqrt{x^{2}-1}\right)}}{\sqrt{x^{2}+1}}+\frac{\ln{\left(x+\sqrt{x^{2}+1}\right)}}{\sqrt{x^{2}-1}}\right]\\ &=z\ln{\left(z+\sqrt{z^{2}+1}\right)}\ln{\left(z+\sqrt{z^{2}-1}\right)}\\ &~~~~~-\int_{1}^{z}\mathrm{d}x\,\frac{x\ln{\left(x+\sqrt{x^{2}-1}\right)}}{\sqrt{x^{2}+1}}-\int_{1}^{z}\mathrm{d}x\,\frac{x\ln{\left(x+\sqrt{x^{2}+1}\right)}}{\sqrt{x^{2}-1}}\\ &=z\ln{\left(z+\sqrt{z^{2}+1}\right)}\ln{\left(z+\sqrt{z^{2}-1}\right)}\\ &~~~~~-\sqrt{z^{2}+1}\ln{\left(z+\sqrt{z^{2}-1}\right)}+\int_{1}^{z}\mathrm{d}x\,\frac{\sqrt{x^{2}+1}}{\sqrt{x^{2}-1}};~~~\small{I.B.P.}\\ &~~~~~-\sqrt{z^{2}-1}\ln{\left(z+\sqrt{z^{2}+1}\right)}+\int_{1}^{z}\mathrm{d}x\,\frac{\sqrt{x^{2}-1}}{\sqrt{x^{2}+1}};~~~\small{I.B.P.}\\ &=z\ln{\left(z+\sqrt{z^{2}+1}\right)}\ln{\left(z+\sqrt{z^{2}-1}\right)}\\ &~~~~~-\sqrt{z^{2}+1}\ln{\left(z+\sqrt{z^{2}-1}\right)}-\sqrt{z^{2}-1}\ln{\left(z+\sqrt{z^{2}+1}\right)}\\ &~~~~~+\int_{1}^{z}\mathrm{d}x\,\left[\frac{\sqrt{x^{2}+1}}{\sqrt{x^{2}-1}}+\frac{\sqrt{x^{2}-1}}{\sqrt{x^{2}+1}}\right]\\ &=z\ln{\left(z+\sqrt{z^{2}+1}\right)}\ln{\left(z+\sqrt{z^{2}-1}\right)}\\ &~~~~~-\sqrt{z^{2}+1}\ln{\left(z+\sqrt{z^{2}-1}\right)}-\sqrt{z^{2}-1}\ln{\left(z+\sqrt{z^{2}+1}\right)}\\ &~~~~~+\int_{1}^{z}\mathrm{d}x\,\frac{2x^{2}}{\sqrt{x^{4}-1}}\\ &=z\operatorname{arsinh}{\left(z\right)}\operatorname{arcosh}{\left(z\right)}-\sqrt{z^{2}+1}\operatorname{arcosh}{\left(z\right)}-\sqrt{z^{2}-1}\operatorname{arsinh}{\left(z\right)}+2\,\mathcal{E}{\left(z\right)},\\ \end{align}$$

where in the last line above we've introduced the auxiliary function $\mathcal{E}:\left(1,\infty\right)\rightarrow\mathbb{R}_{>0}$ to stand for the remaining unsolved integral, i.e.,

$$\mathcal{E}{\left(z\right)}:=\int_{1}^{z}\mathrm{d}x\,\frac{x^{2}}{\sqrt{x^{4}-1}}.$$

Given $z\in\left(1,\infty\right)$,

$$\begin{align} \mathcal{E}{\left(z\right)} &=\int_{1}^{z}\mathrm{d}x\,\frac{x^{2}}{\sqrt{x^{4}-1}}\\ &=\int_{0}^{\frac{\pi}{2}-\arcsin{\left(\frac{1}{z}\right)}}\mathrm{d}\varphi\,\tan{\left(\varphi\right)}\sec{\left(\varphi\right)}\frac{\sec^{2}{\left(\varphi\right)}}{\sqrt{\sec^{4}{\left(\varphi\right)}-1}};~~~\small{\left[x=\sec{\left(\varphi\right)}\right]}\\ &=\int_{0}^{\frac{\pi}{2}-\arcsin{\left(\frac{1}{z}\right)}}\mathrm{d}\varphi\,\frac{\sin{\left(\varphi\right)}}{\cos^{2}{\left(\varphi\right)}\sqrt{1-\cos^{4}{\left(\varphi\right)}}}\\ &=\int_{0}^{\frac{\pi}{2}-\arcsin{\left(\frac{1}{z}\right)}}\mathrm{d}\varphi\,\frac{\sin{\left(\varphi\right)}}{\cos^{2}{\left(\varphi\right)}\sqrt{\left[1-\cos^{2}{\left(\varphi\right)}\right]\left[1+\cos^{2}{\left(\varphi\right)}\right]}}\\ &=\int_{0}^{\frac{\pi}{2}-\arcsin{\left(\frac{1}{z}\right)}}\mathrm{d}\varphi\,\frac{\sin{\left(\varphi\right)}}{\cos^{2}{\left(\varphi\right)}\sqrt{\sin^{2}{\left(\varphi\right)}\left[2-\sin^{2}{\left(\varphi\right)}\right]}}\\ &=\int_{0}^{\frac{\pi}{2}-\arcsin{\left(\frac{1}{z}\right)}}\mathrm{d}\varphi\,\frac{1}{\left[1-\sin^{2}{\left(\varphi\right)}\right]\sqrt{2-\sin^{2}{\left(\varphi\right)}}}\\ &=\int_{0}^{\sin{\left(\frac{\pi}{2}-\arcsin{\left(\frac{1}{z}\right)}\right)}}\mathrm{d}t\,\frac{1}{\left(1-t^{2}\right)\sqrt{1-t^{2}}\sqrt{2-t^{2}}};~~~\small{\left[\varphi=\arcsin{\left(t\right)}\right]}\\ &=\int_{0}^{\cos{\left(\arcsin{\left(\frac{1}{z}\right)}\right)}}\mathrm{d}t\,\frac{1}{\sqrt{\left(1-t^{2}\right)^{3}\left(2-t^{2}\right)}}\\ &=\int_{0}^{\sqrt{1-\sin^{2}{\left(\arcsin{\left(\frac{1}{z}\right)}\right)}}}\mathrm{d}t\,\frac{1}{\sqrt{\left(1-t^{2}\right)^{3}\left(2-t^{2}\right)}}\\ &=\int_{0}^{\sqrt{1-\frac{1}{z^{2}}}}\mathrm{d}t\,\frac{1}{\sqrt{\left(1-t^{2}\right)^{3}\left(2-t^{2}\right)}}.\\ \end{align}$$

Note that $0<\sqrt{1-\frac{1}{z^{2}}}<1$ and set

$$\theta:=\arcsin{\left(\sqrt{1-\frac{1}{z^{2}}}\right)}\land \kappa:=\frac{1}{\sqrt{2}}.$$

The elliptic integral $\mathcal{E}$ can then be reduced to standard form as follows:

$$\begin{align} \mathcal{E}{\left(z\right)} &=\int_{0}^{\sqrt{1-\frac{1}{z^{2}}}}\mathrm{d}t\,\frac{1}{\sqrt{\left(1-t^{2}\right)^{3}\left(2-t^{2}\right)}}\\ &=\int_{0}^{\sqrt{1-\frac{1}{z^{2}}}}\mathrm{d}t\,\frac{2-2t^{2}+t^{4}}{\sqrt{\left(1-t^{2}\right)^{3}\left(2-t^{2}\right)}}-\int_{0}^{\sqrt{1-\frac{1}{z^{2}}}}\mathrm{d}t\,\frac{\left(1-t^{2}\right)^{2}}{\sqrt{\left(1-t^{2}\right)^{3}\left(2-t^{2}\right)}}\\ &=\int_{0}^{\sqrt{1-\frac{1}{z^{2}}}}\mathrm{d}t\,\frac{d}{dt}\left[t\sqrt{\frac{2-t^{2}}{1-t^{2}}}\right]-\int_{0}^{\sqrt{1-\frac{1}{z^{2}}}}\mathrm{d}t\,\frac{\left(1-t^{2}\right)}{\sqrt{\left(1-t^{2}\right)\left(2-t^{2}\right)}}\\ &=\sqrt{1-\frac{1}{z^{2}}}\sqrt{z^{2}+1}-\int_{0}^{\sin{\left(\theta\right)}}\mathrm{d}t\,\frac{\left(1-t^{2}\right)}{\sqrt{\left(1-t^{2}\right)\left(2-t^{2}\right)}}\\ &=\frac{\sqrt{z^{4}-1}}{z}+\int_{0}^{\sin{\left(\theta\right)}}\mathrm{d}t\,\frac{1-\left(2-t^{2}\right)}{\sqrt{\left(1-t^{2}\right)\left(2-t^{2}\right)}}\\ &=\frac{\sqrt{z^{4}-1}}{z}+\int_{0}^{\sin{\left(\theta\right)}}\mathrm{d}t\,\frac{1}{\sqrt{\left(1-t^{2}\right)\left(2-t^{2}\right)}}-\int_{0}^{\sin{\left(\theta\right)}}\mathrm{d}t\,\frac{\sqrt{2-t^{2}}}{\sqrt{1-t^{2}}}\\ &=\frac{\sqrt{z^{4}-1}}{z}+\frac{1}{\sqrt{2}}\int_{0}^{\sin{\left(\theta\right)}}\mathrm{d}t\,\frac{1}{\sqrt{\left(1-t^{2}\right)\left(1-\frac12t^{2}\right)}}-\sqrt{2}\int_{0}^{\sin{\left(\theta\right)}}\mathrm{d}t\,\frac{\sqrt{1-\frac12t^{2}}}{\sqrt{1-t^{2}}}\\ &=\frac{\sqrt{z^{4}-1}}{z}+\frac{1}{\sqrt{2}}\int_{0}^{\sin{\left(\theta\right)}}\mathrm{d}t\,\frac{1}{\sqrt{\left(1-t^{2}\right)\left(1-\kappa^{2}t^{2}\right)}}-\sqrt{2}\int_{0}^{\sin{\left(\theta\right)}}\mathrm{d}t\,\frac{\sqrt{1-\kappa^{2}t^{2}}}{\sqrt{1-t^{2}}}\\ &=\frac{\sqrt{z^{4}-1}}{z}+\frac{1}{\sqrt{2}}F{\left(\theta,\kappa\right)}-\sqrt{2}E{\left(\theta,\kappa\right)}.\\ \end{align}$$

Putting everything together, our final result is

$$\begin{align} \mathcal{I}{\left(z\right)} &=\int_{1}^{z}\mathrm{d}x\,\ln{\left(x+\sqrt{x^{2}+1}\right)}\ln{\left(x+\sqrt{x^{2}-1}\right)}\\ &=z\operatorname{arsinh}{\left(z\right)}\operatorname{arcosh}{\left(z\right)}-\sqrt{z^{2}+1}\operatorname{arcosh}{\left(z\right)}-\sqrt{z^{2}-1}\operatorname{arsinh}{\left(z\right)}\\ &~~~~~+\frac{2\sqrt{z^{4}-1}}{z}+\sqrt{2}F{\left(\theta,\kappa\right)}-2\sqrt{2}E{\left(\theta,\kappa\right)}.\blacksquare\\ \end{align}$$

David H
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