Evaluating $\int_1^{\sqrt2} \frac{\tanh^{-1}(\sqrt{2-x^2})}{1+x} dx$

Question

I was trying to compute the value of the integral

$$\ I = \int_1^{\sqrt2} \frac{\operatorname{arctanh}(\sqrt{2-x^2})}{1+x}dx$$

I began by declaring the family of integrals:

$$\ I(a) = \int_1^{\sqrt2} \frac{\operatorname{arctanh}(a\sqrt{2-x^2})}{1+x}dx$$

And then differentiating under the integral sign in accordance to the Leibnitz rule, keeping in mind that:

$$\frac{d}{dx}\operatorname{arctanh}(x) = \frac{1}{1-x^2}$$

I obtained:

$$\ I'(a) = \int_1^{\sqrt2} \frac{\sqrt{2-x^2}dx}{(1-a^2(2-x^2))\cdot (1+x)}$$

Now I substitute $\ x = \sqrt2 sin\theta$. Changing the bounds, I received:

$$\ I'(a) = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{2\cos^2\theta d\theta}{(1-2a^2\cos^2\theta)\cdot (1+\sqrt2 \sin\theta)}$$

The problem with this integral is that - When the cosine term in the denominator varies from $\frac{1}{\sqrt2}$ to $0$, it will reach a point within the above range such that

$$\ 1 = 2a^2cos^2\theta$$

For all non zero $a>1$, it will have an asymptote. I do not understand how we will be able to then simplify the integral further. Maybe the radius of convergence of the integral necessitates that $\ a<1$?

If we assume that the integral converges, I think we can evaluate $I'(a)$ by multiplying and dividing by $\frac{-1}{a^2}$.

I am getting a not-so-good expression for $I'(a)$:

$$\ I'(a) = \frac{-1}{a^2}\cdot \big[ \frac{1}{2}\cdot \log(2) - \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{d\theta}{(1-2a^2\cos^2\theta)(1+\sqrt2\sin\theta)}\big]$$

Note that for the special case $\ a = 1$, it is probably the case that

$$\ I(1) = \frac{\pi^2}{48}$$

This was deduced from WolframAlpha as a possible closed form for the integral. I have not confirmed whether this is the real answer or not, but it does seem to match.

Any help is appreciated. I am also open to other techniques for evaluating the original integral. Thanks for reading!

Answer 1

Substitute $x=\sqrt2\,\sin y$, exploit symmetry, convert from hyperbolic to logarithmic, and substitute $y=2\arctan z$:

$$\begin{align*} I &= \int_1^{\sqrt2} \frac{\operatorname{artanh} \sqrt{2-x^2}}{1+x} \, dx \\ &= \sqrt2 \int_{\tfrac\pi4}^{\tfrac\pi2} \frac{\cos y \operatorname{artanh}\left(\sqrt2\,\cos y\right)}{1+\sqrt2\,\sin y} \, dy \\ &= \sqrt2 \int_0^{\tfrac\pi4} \frac{\cos \left(y+\frac\pi4\right) \operatorname{artanh}\left(\sqrt2\,\cos \left(y+\frac\pi4\right)\right)}{1+\sqrt2\,\sin \left(y+\frac\pi4\right)} \, d\left(y+\frac\pi4\right) \\ &= \int_0^{\tfrac\pi4} \frac{(\cos y-\sin y) \operatorname{artanh}\left(\cos y-\sin y\right)}{1+\cos y+\sin y} \, dy \tag{$*$} \\ &= \int_{\tfrac\pi2}^{\tfrac\pi4} \frac{\left(\cos\left(\frac\pi2-y\right)-\sin\left(\frac\pi2-y\right)\right) \operatorname{artanh}\left(\cos \left(\frac\pi2-y\right)-\sin \left(\frac\pi2-y\right)\right)}{1+\cos\left(\frac\pi2-y\right)+\sin \left(\frac\pi2-y\right)} \, d\left(\frac\pi2-y\right) \\ &= \int_{\tfrac\pi4}^{\tfrac\pi2} \frac{(\cos y-\sin y) \operatorname{artanh}\left(\cos y-\sin y\right)}{1+\cos y+\sin y} \, dy \tag{$**$} \\ &= \frac14 \int_0^{\tfrac\pi2} \frac{\cos y - \sin y}{1+\cos y+\sin y} \log \frac{1+\cos y-\sin y}{1-\cos y+\sin y} \, dy \tag*{$\frac{(*)+(**)}2$}\\ &= \frac14 \int_0^1 \frac{1-2z-z^2}{(1+z)\left(1+z^2\right)} \log \frac{1-z}{z(1+z)} \, dz \end{align*}$$

Expanding into partial fractions and breaking up the logarithms leads to integrals of the much simpler forms,

$$\int_0^1 \frac{L(z)}{1+z}\,dz \text{ and } \int_0^1 \frac{2z L(z)}{1+z^2} \, dz$$

where $L(z) \in \left\{\log z, \log(1\pm z)\right\}$. Closed forms are listed below. Linked Q&As were found with Approach Zero.

$$\begin{align*} \int_0^1 \frac{\log z}{1+z} \, dz &= -\frac{\pi^2}{12} \tag1 \\ \int_0^1 \frac{\log(1-z)}{1+z} \, dz &= \frac12 \log^22 - \frac{\pi^2}{12} \tag2 \\ \int_0^1 \frac{\log(1+z)}{1+z} \, dz &= \frac12 \log^22 \\[2ex] \int_0^1 \frac{2z \log z}{1+z^2} \, dz &= -\frac{\pi^2}{24} \tag3 \\ \int_0^1 \frac{2z \log(1-z)}{1+z^2} \, dz &= \frac14 \log^22 - \frac{5\pi^2}{48} \tag4 \\ \int_0^1 \frac{2z \log(1+z)}{1+z^2} \, dz &= \frac14 \log^22 + \frac{\pi^2}{48} \tag4 \end{align*}$$

$(1)$, $(2)$, $(3)$, $(4)$

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On the other hand, as @Zacky pointed out, by replacing $z\mapsto\dfrac{1-z}{1+z}$ one can collapse $I$ into a multiple of integral $(1)$ :

$$\begin{align*} \int_0^1 \frac{\log \frac{1-z}{1+z}}{1+z} \, dz &= \int_0^1 \frac{\log z}{1+z} \, dz \\[2ex] \int_0^1 \frac{2z \log \frac{1-z}{1+z}}{1+z^2} \, dz &= \int_0^1 \frac{2\log z}{1+z} \, dz - \int_0^1 \frac{2z \log z}{1+z^2} \, dz \\ &= \int_0^1 \frac{2\log z}{1+z} \, dz - \int_0^1 \frac{\frac12 \log z^2}{1+z^2} \, dz^2 \\ &= \frac32 \int_0^1 \frac{\log z}{1+z} \, dz \end{align*}$$