Show $\int_0^{\frac1{\sqrt3}} \frac{\cot^{-1}\sqrt{2-x^2}}{1+x^2}dx=\frac{\pi^2}{30}$

Question

I am unable to show $$\int_0^{\frac1{\sqrt3}} \frac{\cot^{-1}\sqrt{2-x^2}}{1+x^2}dx=\frac{\pi^2}{30} $$ except with a few observations below.

1). Despite the appearance, I do not believe this integral is related to the Ahmed integral or any variations of known solution.

2). The integral is related to a more complex one posted here. However, the answer offered is non conventional, which I could not fully appreciate.

3). Numerically, I could establish that $$\int_0^{\frac1{\sqrt3}} \frac{\cot^{-1}\sqrt{2-x^2}}{1+x^2}dx =2 \int_0^{\frac1{\sqrt3}} \frac{\cot^{-1}\sqrt5 x}{1+x^2}dx - \int_0^{\sqrt3} \frac{\cot^{-1}\sqrt5 x}{1+x^2}dx $$ where the RHS can be worked out with an elaborate procedure.

So, I would like to see a traditional solution without special functions, or, at least, establish the integral relationship above analytically.

Answer 1

Perform the substitution $t=\frac{1}{\sqrt{2-x^2}}$ to check that, $$ \tag{1}\int_{0}^{\frac{1}{\sqrt{3} } } \frac{\operatorname{arccot} \left ( \sqrt{2-x^2} \right ) }{ 1+x^2}\text{d}x= \int_{\frac{\sqrt{2}}{2} }^{\sqrt{\frac{3}{5}}} \frac{\arctan\left ( t \right ) }{\sqrt{2t^2-1}(3t^2-1) }\text{d}t. $$ The second integral can be done by evaluating $$ \tag{2}\int_{0}^{1}\int_{0}^{1}\frac{1}{(1+x^2)(1+y^2)\sqrt{3+x^2+y^2} }\text{d}x\text{d}y =\frac{\pi^2}{30}. $$ To prove $(2)$, use the identity $$ \int_{0}^{\infty} {e^{-t^2x^2}}\text{d}x =\frac{\sqrt{\pi} }{2\sqrt{t} }. $$ Then I can get (classical result $\displaystyle{\int_{0}^{1} \frac{e^{-t^2x^2}}{1+x^2}\text{d}x =\frac{\pi}{4}e^{t^2}\left ( 1-\operatorname{erf}(t)^2 \right )}$ is used) $$ \begin{align*} &\int_{0}^{1}\int_{0}^{1}\frac{1}{(1+x^2)(1+y^2)\sqrt{3+x^2+y^2} }\text{d}x\text{d}y\\ =& \frac{2}{\sqrt{\pi} } \int_{0}^{1} \int_{0}^{1} \frac{1}{(1+x^2)(1+y^2)} \left (\int_{0}^{\infty}e^{-t^2(3+x^2+y^2)}\text{d}t\right ) \text{d}x\text{d}y\\ \overset{{\scriptsize\text{Fubini}}}{=}& \frac{2}{\sqrt{\pi} } \int_{0}^{\infty}e^{-3t^2}\left ( \int_{0}^{1} \frac{e^{-t^2x^2}}{1+x^2}\text{d} x\right )^2\text{d}t\\ =&\frac{2}{\sqrt{\pi} }\cdot\frac{\pi^2}{16} \int_{0}^{\infty} e^{-x^2}\left(1-\operatorname{erf}(x)^2\right)^2\text{d}x\\ =&\frac{2}{\sqrt{\pi} }\cdot\frac{\pi^2}{16} \int_{0}^{\infty} e^{-x^2}\left(1-2\operatorname{erf}(x)^2+\operatorname{erf}(x)^4\right)\text{d}x\\ =&\frac{2}{\sqrt{\pi} }\cdot\frac{\pi^2}{16}\left(\int_{0}^{\infty}e^{-x^2}\text{d}x -2\int_{0}^{\infty}e^{-x^2}\operatorname{erf}(x)^2\text{d}x +\int_{0}^{\infty}e^{-x^2}\operatorname{erf}(x)^4\text{d}x\right)\\ =&\frac{\pi^2}{30}. \end{align*} $$ The last equality follows from their primitives $$ \frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{\sqrt{\pi}}{2} \operatorname{erf}(x)^{n+1} \right ) =(n+1)e^{-t^2}\operatorname{erf}(x)^{n}. $$ And $$ \begin{align*} &\int_{0}^{1}\int_{0}^{1}\frac{1}{(1+x^2)(1+y^2)\sqrt{3+x^2+y^2} }\text{d}x\text{d}y\\ =&\int_{0}^{1} \frac{\arctan\left ( \sqrt{\frac{2+x^2}{4+x^2} } \right ) }{ \left ( 1+x^2 \right )\sqrt{2+x^2} }\text{d}x \\ \overset{{\scriptsize t=\sqrt{\frac{2+x^2}{4+x^2} }}}{=}&\int_{\frac{\sqrt{2}}{2} }^{\sqrt{\frac{3}{5}}} \frac{\arctan\left ( t \right ) }{\sqrt{2t^2-1}(3t^2-1) }\text{d}t\\ =&\int_{0}^{\frac{1}{\sqrt{3} } } \frac{\operatorname{arccot} \left ( \sqrt{2-x^2} \right ) }{ 1+x^2}\text{d}x=\frac{\pi^2}{30}. \end{align*} $$

Answer 2

To establish the integral relationship we can proceed as with this integral.

$$\int_0^\frac{1}{\sqrt 3}\frac{\operatorname{arccot}\left(\sqrt{2-x^2}\right)}{1+x^2}dx\overset{x\to \frac{1}{\sqrt x}} = \frac12\int_3^\infty \frac{\arctan\left(\frac{\sqrt x}{\sqrt{2x-1}}\right)}{\sqrt x(1+x)}dx$$

$$=\frac12\int_3^\infty \frac{1}{1+x}\left(\int_0^{\large \frac{1}{\sqrt{2x-1}}} \frac{1}{1+xy^2}dy\right)dx $$

$$= \frac12\int_0^\frac{1}{\sqrt 5}\int_3^{\large \frac{1+y^2}{2y^2}}\frac{dxdy}{(1+x)(1+y^2x)}=\frac12\int_0^\frac{1}{\sqrt 5} \frac{\ln\left(\frac{(1+3y^2)^2}{4y^2 (3+y^2)}\right)}{1-y^2}dy \tag{*}$$

We also have:

$$\int_0^\frac{1}{\sqrt 3} \frac{\operatorname{arccot}(\sqrt 5 x)}{1+x^2}dx=\int_0^\frac{1}{\sqrt 5}\int_0^\frac{1}{\sqrt 3}\frac{x}{(y^2+x^2)(1+x^2)}dxdy=\frac12 \int_0^\frac{1}{\sqrt 5} \frac{\ln\left(\frac{1+3y^2}{4y^2}\right)}{1-y^2}dy$$

$$\int_0^{\sqrt 3} \frac{\operatorname{arccot}(\sqrt 5 x)}{1+x^2}dx=\int_0^\frac{1}{\sqrt 5} \int_0^{\sqrt 3}\frac{x}{(y^2+x^2)(1+x^2)}dxdy=\frac12\int_0^\frac{1}{\sqrt 5} \frac{\ln\left(\frac{3+y^2}{4y^2}\right)}{1-y^2}dy$$

$$\Rightarrow \boxed{\int_0^\frac{1}{\sqrt 3}\frac{\operatorname{arccot}\left(\sqrt{2-x^2}\right)}{1+x^2}dx=2\int_0^\frac{1}{\sqrt 3} \frac{\operatorname{arccot}(\sqrt 5 x)}{1+x^2}dx-\int_0^{\sqrt 3} \frac{\operatorname{arccot}(\sqrt 5 x)}{1+x^2}dx}$$

---

It should also be mentioned that continuing from $(*)$, we get:

$$\frac12\int_0^\frac{1}{\sqrt 5} \frac{\ln\left(\frac{(1+3y^2)^2}{4y^2 (3+y^2)}\right)}{1-y^2}dy\overset{y=\frac{1-x}{1+x}}=\frac14\int_\frac{1}{\phi^2}^1\frac{\ln\left(\frac{(1+x^3)^2}{(1-x^3)(1-x)(1+x)^2}\right)}{x}dx,\ \phi=\frac{1+\sqrt 5}{2}$$

$$=\frac{\pi^2}{90} +\frac14 \ln^2(\phi)+\frac16\operatorname{Li_2}\left(-\frac{1}{\phi^6}\right)-\frac{1}{12}\operatorname{Li_2}\left(\frac{1}{\phi^6}\right)-\frac12\operatorname{Li_2}\left(-\frac{1}{\phi^2}\right)$$

And the remaining linear combination of dilogarithms might be extracted as it was described here, providing the final part of $\frac{\pi^2}{30}$, but that'll be just a brute force approach. Instead maybe it's more elegant to derive, combined with the other answer, the following dilogarithm identity (which might be unknown):

$$\boxed{2\operatorname{Li}_2\left(-\phi^{-6}\right)-\operatorname{Li}_2\left(\phi^{-6}\right)-6\operatorname{Li}_2\left(-\phi^{-2}\right)=\frac{4\pi^2}{15}-3\ln^2\left(\phi\right)}$$