Proving $\int_{\sqrt{\frac{3}{5}}}^1 \frac{\arctan (x)}{\sqrt{2 x^2-1} \left(3 x^2-1\right)} \, dx=\frac{3 \pi ^2}{160}$

Question

How to prove $$\int_{\sqrt{3/5}}^1 \frac{\arctan (x)}{\sqrt{2 x^2-1} \left(3 x^2-1\right)} \, dx=\frac{3\pi^2}{160}$$ I found the integral neat enough but also tough. Is it somehow related to the Ahmed integral$?$ Any help will be appreciated.

---

Update: Please see the link under @pisco's answer for further reference.

Answer 1

Two proofs will be given. One proof is the remaining of this answer, along lines of Schläfli and Coxeter. Second proof, more direct, is given at this answer's remark.

---

Note that $$\int_{\sqrt {3/5} }^1 {\frac{{\arctan x}}{{\sqrt {2{x^2} - 1} (3{x^2} - 1)}}dx} = \int_1^{\sqrt {5/3} } {\frac{{x(\frac{\pi }{2} - \arctan x)}}{{\sqrt {2 - {x^2}} (3 - {x^2})}}dx} = \frac{{{\pi ^2}}}{{24}} - \frac{1}{2}\int_1^{5/3} {\frac{{\arctan \sqrt x }}{{\sqrt {2 - x} (3 - x)}}dx}$$ therefore your integral is equivalent to $$\int_1^{5/3} {\frac{{\arctan \sqrt x }}{{\sqrt {2 - x} (3 - x)}}dx} = \frac{{11{\pi ^2}}}{{240}}$$ I will prove this by establishing $$\tag{1}\int_0^1 {\frac{{{{\tan }^{ - 1}}\sqrt t }}{{\sqrt {2 - t} (3 - t)}}dt} = \frac{{{\pi ^2}}}{{48}}$$ $$\tag{2} \int_0^{5/3} {\frac{{{{\tan }^{ - 1}}\sqrt t }}{{\sqrt {2 - t} (3 - t)}}dt} = \frac{{{\pi ^2}}}{{15}}$$

---

The following ideas are an expounded version of Polylogarihm and Associated Functions by Leonard Lewin page 115-117, arguments there are very terse.

Let $$S(\alpha ,\beta ,\gamma ) = \sum\limits_{n = 1}^\infty {\frac{{{k^n}}}{{{n^2}}}(\cos 2n\alpha - \cos 2n\beta + \cos 2n\gamma - 1)} - {\alpha ^2} + {\beta ^2} - {\gamma ^2}$$ where $0\leq \alpha,\gamma \leq \pi/2, 0\leq \beta < \pi$ and $k$ is a function of $\alpha,\beta,\gamma$ which will be determined soon. It is easy to check that $$\frac{{\partial S}}{{\partial \alpha }} = - 2\sum\limits_{n = 1}^\infty {\frac{{{k^n}}}{n}\sin 2n\alpha } - 2\alpha = - 2{\tan ^{ - 1}}\left( {\frac{{1 + k}}{{1 - k}}\tan \alpha } \right)$$ Therefore the differential $dS$ is $$\begin{aligned}dS &= \frac{1}{k}\sum\limits_{n = 1}^\infty {\frac{{{k^n}}}{n}(\cos 2n\alpha - \cos 2n\beta + \cos 2n\gamma - 1)} dk - 2{\tan ^{ - 1}}\left( {\frac{{1 + k}}{{1 - k}}\tan \alpha } \right)d\alpha+\\ & 2{\tan ^{ - 1}}\left( {\frac{{1 + k}}{{1 - k}}\tan \beta } \right)d\beta - 2{\tan ^{ - 1}}\left( {\frac{{1 + k}}{{1 - k}}\tan \gamma } \right)d\gamma \end{aligned}$$ Now choose $k$ such that the coefficient of $dk$ vanishes, using $\sum\limits_{n = 1}^\infty {\frac{{{a^n}\cos nx}}{n}} = - \frac{1}{2}\ln ({a^2} - 2a\cos x + 1)$ one easily see such $k$ is $$ k = \frac{{\sqrt {{{\cos }^2}\alpha {{\cos }^2}\gamma - {{\cos }^2}\beta } - \sin \alpha \sin \gamma }}{{\sqrt {{{\cos }^2}\alpha {{\cos }^2}\gamma - {{\cos }^2}\beta } + \sin \alpha \sin \gamma }}$$ This completes the definition of $S(\alpha,\beta,\gamma)$. Note that in order for $k$ to be real, we need to assume the term inside radical is always $\geq 0$, we confine ourselves exclusive to this case. Now $dS$ becomes $$\tag{3}dS = - 2{\tan ^{ - 1}}\left( {\frac{{1 + k}}{{1 - k}}\tan \alpha } \right)d\alpha+ 2{\tan ^{ - 1}}\left( {\frac{{1 + k}}{{1 - k}}\tan \beta } \right)d\beta - 2{\tan ^{ - 1}}\left( {\frac{{1 + k}}{{1 - k}}\tan \gamma } \right)d\gamma $$

Four observations:

• $S(0,\beta,\gamma) = \pi(\beta-\gamma)$
• When $\sin^2 \alpha + \sin^2 \gamma = \sin^2 \beta$, $S(\alpha,\beta,\gamma) = -\alpha^2+\beta^2-\gamma^2$
• When $\cos \alpha \cos\gamma = \cos\beta$, $S(\alpha,\beta,\gamma)= 0$.
• $S(\alpha,\pi - 2\alpha,\alpha) = 6S(\alpha,\pi/3,\pi/6)$

Proof. For the first one, $\alpha = 0$ implies $k=1$, so $$S(0,\beta ,\gamma ) = \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}( - \cos 2n\beta + \cos 2n\gamma )} + {\beta ^2} - {\gamma ^2} = \pi (\beta - \gamma )$$ For second one, the condition implies $k=0$. For third one, the condition implies $k=-1$, then just evaluate the series. The fourth assertion is more crucial. We prove it carefully. By first bullet, it suffices to prove their derivative with respect to $\alpha$ are equal. When $\beta = \pi/3, \gamma = \pi/6$, one sees that $$\frac{{1 + k}}{{1 - k}}\tan \alpha = \sqrt{3 \cos^2 \alpha-1} \sec \alpha $$ Since $\beta,\gamma$ are constants, $d\gamma = d\beta = 0$, $(3)$ shows $$\tag{A}-\frac{1}{2}\frac{dS(\alpha,\pi/3,\pi/6)}{d\alpha} = \arctan(\sqrt{3 \cos^2 \alpha-1} \sec \alpha)$$

Now consider $S(\alpha,\pi - 2\alpha,\alpha)$, with $\beta = \pi - 2\alpha, \gamma = \alpha, d\beta = -2d\alpha$, $d\gamma = d\alpha$, one computes via $(3)$, $$\tag{B}-\frac{1}{2}\frac{dS(\alpha,\pi - 2\alpha,\alpha)}{d\alpha} = 2 \arctan \left(\sqrt{\cos ^4\alpha-\cos ^2 2\alpha} \csc \alpha \sec \alpha\right)-2 \arctan \left(\sqrt{\cos ^4 \alpha-\cos ^2 2 \alpha} \tan 2 \alpha \csc ^2 \alpha \right)+2\pi $$ To complete the proof, it suffices to differentiate RHS of $6\times (A), (B)$, and see whether they are equal. This becomes a trivial but computational heavy task.

---

Now let $a>b$, consider $$\sqrt {a - b} \int_0^x {\frac{{{{\tan }^{ - 1}}\sqrt t }}{{\sqrt {b - t} (a - t)}}dt} = - 2\int_0^x {{{\tan }^{ - 1}}\sqrt t d({{\tan }^{ - 1}}\sqrt {\frac{{b - t}}{{a - b}}} )} $$ fix $\alpha = {\tan ^{ - 1}}\sqrt {\frac{{b - t}}{{a - b}}}$, we find $\beta,\gamma$ (depends on $a,b$ but not on $t$) such that $$\frac{{1 + k}}{{1 - k}}\tan \alpha = \sqrt t $$ One easily verifies one such pair $\beta,\gamma$ is $$\gamma = {\tan ^{ - 1}}\frac{1}{{\sqrt a }}\qquad \beta = {\tan ^{ - 1}}\sqrt {\frac{{b + 1}}{{a - b}}} $$ Since $\beta,\gamma$ are independent of $t$, $(3)$ implies $$dS(\alpha ,{\tan ^{ - 1}}\sqrt {\frac{{b + 1}}{{a - b}}} ,{\tan ^{ - 1}}\frac{1}{{\sqrt a }}) = - 2{\tan ^{ - 1}}\left( {\frac{{1 + k}}{{1 - k}}\tan \alpha } \right)d\alpha = - 2{\tan ^{ - 1}}\sqrt t d\alpha $$ Hence $$\int_0^x { - 2{{\tan }^{ - 1}}\sqrt t d\alpha } = S({\tan ^{ - 1}}\sqrt {\frac{{b - x}}{{a - b}}} ,{\tan ^{ - 1}}\sqrt {\frac{{b + 1}}{{a - b}}} ,{\tan ^{ - 1}}\frac{1}{{\sqrt a }}) + C$$ for a constant $C$ independent of $x$. Set $x=0$, then one checks the third bullet point applies for $$S({\tan ^{ - 1}}\sqrt {\frac{b}{{a - b}}} ,{\tan ^{ - 1}}\sqrt {\frac{{b + 1}}{{a - b}}} ,{\tan ^{ - 1}}\frac{1}{{\sqrt a }})$$ therefore it is $0$, hence $C=0$. Thus we proved

$$\tag{4}\int_0^x {\frac{{{{\tan }^{ - 1}}\sqrt t }}{{(a - t)\sqrt {b - t} }}dt} = \frac{1}{{\sqrt {a - b} }}S({\tan ^{ - 1}}\sqrt {\frac{{b - x}}{{a - b}}} ,{\tan ^{ - 1}}\sqrt {\frac{{b + 1}}{{a - b}}} ,{\tan ^{ - 1}}\frac{1}{{\sqrt a }})$$

---

Now let $a=3, b=2$, we deduce from $(4)$ $$\int_0^1 {\frac{{{{\tan }^{ - 1}}\sqrt t }}{{\sqrt {2 - t} (3 - t)}}dt} = S(\frac{\pi }{4},\frac{\pi }{3},\frac{\pi }{6})\qquad \int_0^{5/3} {\frac{{{{\tan }^{ - 1}}\sqrt t }}{{\sqrt {2 - t} (3 - t)}}dt} = S(\frac{\pi }{6},\frac{\pi }{3},\frac{\pi }{6})$$ Now the second bullet applies for the former one, so $S(\frac{\pi }{4},\frac{\pi }{3},\frac{\pi }{6}) = \frac{{{\pi ^2}}}{{48}}$, this is $(1)$. For latter one, note that fourth bullet implies $$6S(\frac{\pi }{6},\frac{\pi }{3},\frac{\pi }{6}) = S(\frac{\pi }{6},\frac{{2\pi }}{3},\frac{\pi }{6})$$ but directly from definition (the $k$ associated with these two pairs are equal), one sees that $$S(\frac{\pi }{6},\frac{{2\pi }}{3},\frac{\pi }{6}) - S(\frac{\pi }{6},\frac{\pi }{3},\frac{\pi }{6}) = \sum\limits_{n = 1}^\infty {\frac{{{k^n}}}{{{n^2}}}(\underbrace{\cos \frac{{2\pi n}}{3} - \cos \frac{{4\pi n}}{3}}_{=0})} + {(\frac{{2\pi }}{3})^2} - {(\frac{\pi }{3})^2} = \frac{\pi^2}{3}$$ thus $S(\frac{\pi }{6},\frac{\pi }{3},\frac{\pi }{6}) = \frac{{{\pi ^2}}}{{15}}$, this is $(2)$. The integral claimed by OP is now established.

Answer 2

This is a partial solution which transformed the original integral into an Ahmed-like integral. Firstly, substitute $y\to \sqrt{2 x^2-1}$ and introduce a parameter $a$ into $\tan ^{-1}\left(a \sqrt{\frac{1}{2} \left(x^2+1\right)}\right)$. Differentiate w.r.t $a$, the integrand will be a rational function. After integration w.r.t $y$, we have $I=-A+B+\frac{C \left(\pi -3 \tan ^{-1}\left(\sqrt{\frac{3}{5}}\right)\right)}{\sqrt{3}}$ where $\small A=\int_0^1 \frac{x \tan ^{-1}\left(\frac{x}{\sqrt{x^2+2}}\right)}{\sqrt{x^2+2} \left(x^2+3\right)} \, dx=\frac{\pi ^2}{288},$ $\small B=\int_0^1 \frac{x \tan ^{-1}\left(\frac{x}{\sqrt{5} \sqrt{x^2+2}}\right)}{\sqrt{x^2+2} \left(x^2+3\right)} \, dx,$ $ C=\int_0^1 \frac{1}{x^2+3} \, dx=\frac{\pi }{6 \sqrt{3}}$. For $A$, integrate by parts using $\small \int \frac{x}{\sqrt{x^2+2} \left(x^2+3\right)} \, dx=\tan ^{-1}\left(\sqrt{x^2+2}\right)$ it reduces to the original Ahmed integral. Apply the same method to $B$, I arrive at corresponding $\small B'=\int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+2}\right)}{\sqrt{x^2+2} \left(3 x^2+5\right)} \, dx$ whose value should be $\small\frac{1}{5} \sqrt{5} \left(\frac{\pi ^2}{30}-\frac{1}{6} \pi \tan ^{-1}\left(\sqrt{\frac{3}{5}}\right)+\frac{1}{3} \pi \tan ^{-1}\left(\sqrt{\frac{1}{15}}\right)\right)$ due to the conjectural result. So now we only need to justify the value of $B'$.

---

Update: According to the link FDP provided under pisco's answer, we are able to evaluate more nontrivial Coxeter integrals such as $\int_0^{\frac{\pi }{5}} \cos ^{-1}\left(\frac{\cos (x)}{2 \cos (x)+1}\right) \, dx=\frac{71 \pi ^2}{900}$. Based on $S$ function's representation as well as Feynman's trick, an elegant formula is found:

• $\small \int_{\frac{1}{\sqrt{y}}}^{\frac{1}{\sqrt{x}}} \frac{\arctan (x)}{\sqrt{2 x^2-1} \left(3 x^2-1\right)} \, dx=\frac{1}{2} \pi \left(\tan ^{-1}\left(\sqrt{2-y}\right)-\tan ^{-1}\left(\sqrt{2-x}\right)\right)+\frac{1}{2}\left(S\left(\tan ^{-1}\left(\sqrt{2-y}\right),\frac{\pi }{3},\frac{\pi }{6}\right)-S\left(\tan ^{-1}\left(\sqrt{2-x}\right),\frac{\pi }{3},\frac{\pi }{6}\right)\right)=\left(\sqrt{y (2-y)} A(y)-\sqrt{x (2-x)} A(x)\right)+\frac{1}{6} \pi \left(\tan ^{-1}\left(\sqrt{\frac{3 (2-y)}{y}}\right)-\tan ^{-1}\left(\sqrt{\frac{3 (2-x)}{x}}\right)\right)-\frac{1}{3} \pi \left(\tan ^{-1}\left(\sqrt{\frac{2-y}{3 y}}\right)-\tan ^{-1}\left(\sqrt{\frac{2-x}{3 x}}\right)\right)$

Where function $S$ is the one defined in pisco's answer, $A$ the generalized Ahmed integral:

• $A(t)=\int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+2}\right)}{\sqrt{x^2+2} \left(t+x^2\right)} \, dx,A(1)=\frac{5 \pi ^2}{96}$

Using the original Ahmed integral and special value of $S$, one may let $x\to1$ and assign special values to $y$ to evaluate, say

• $\small \int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+2}\right)}{\sqrt{x^2+2} \left(x^2+\frac{2 \sqrt{5}}{5}+1\right)} \, dx=\sqrt{5} \left(\frac{71 \pi ^2}{3600}+\frac{1}{3} \pi \tan ^{-1}\left(\sqrt{\frac{1}{3} \left(9-4 \sqrt{5}\right)}\right)-\frac{1}{6} \pi \tan ^{-1}\left(\sqrt{27-12 \sqrt{5}}\right)\right)$
• $\small \int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+2}\right)}{\sqrt{x^2+2} \left(x^2-\frac{2 \sqrt{5}}{5}+1\right)} \, dx=\sqrt{5} \left(\frac{241 \pi ^2}{3600}+\frac{1}{3} \pi \tan ^{-1}\left(\sqrt{\frac{1}{3} \left(4 \sqrt{5}+9\right)}\right)-\frac{1}{6} \pi \tan ^{-1}\left(\sqrt{12 \sqrt{5}+27}\right)\right)$

From which we deduce the final one (via PFD), a remarkable quartic Ahmed integral:

• $\int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+2}\right)}{\sqrt{x^2+2} \left(5 x^4+10 x^2+1\right)} \, dx=\frac{37 \pi ^2}{1440}$

Answer 3

Some substitutions which I think make the integral more simple.

$$I=\int_{\sqrt{\frac{3}{5}}}^1 \frac{\arctan (x)}{\sqrt{2 x^2-1} \left(3 x^2-1\right)} \, dx$$

$$x=\frac{1}{y}$$

$$I=\int_1^{\sqrt{\frac{5}{3}}} \frac{y \arctan \frac{1}{y}}{\sqrt{2 -y^2} \left(3-y^2\right)} \, dy$$

$$y=\sqrt{2} z$$

$$I=\frac{\sqrt{2}}{3} \int_{\frac{1}{\sqrt{2}}}^{\sqrt{\frac{5}{6}}} \frac{z \arctan \frac{1}{\sqrt{2} z}}{\sqrt{1 -z^2} \left(1-\frac23 z^2\right)} \, dz$$

$$z^2=u$$

$$I=\frac{\sqrt{2}}{6} \int_{\frac{1}{2}}^{\frac{5}{6}} \frac{\arctan \frac{1}{\sqrt{2 u} }}{\sqrt{1 -u} \left(1-\frac23 u\right)} \, du$$

$$u= \frac{1+s}{2}$$

$$I=\frac{1}{4} \int_{0}^{\frac{2}{3}} \frac{\arctan \frac{1}{\sqrt{1+s} }}{\sqrt{1 -s} \left(1-\frac12 s\right)} \, ds$$

So we need to prove that:

$$J=\int_{0}^{\frac{2}{3}} \frac{\arctan \frac{1}{\sqrt{1+s} }}{\sqrt{1 -s} \left(1-\frac12 s\right)} \, ds= \frac{3 \pi^2}{40}$$

Let's try integration by parts. Turns out that:

$$ \int \frac{ds}{\sqrt{1 -s} \left(1-\frac12 s\right)}=-4 \arctan \sqrt{1-s}$$

$$\frac{d}{ds} \arctan \frac{1}{\sqrt{1+s} }=-\frac{1}{4} \frac{ds}{\sqrt{1 +s} \left(1+\frac12 s\right)}$$

So our integral is equal to:

$$J=-4 \arctan\frac{1}{\sqrt{1+s} } \arctan \sqrt{1-s} \bigg|_0^{2/3}-\int_{0}^{\frac{2}{3}} \frac{\arctan \sqrt{1-s}}{\sqrt{1 +s} \left(1+\frac12 s\right)} \, ds$$

$$J=\frac{\pi^2}{4}-\frac{2 \pi}{3} \arctan \sqrt{\frac{3}{5}} -\int_{0}^{\frac{2}{3}} \frac{\arctan \sqrt{1-s}}{\sqrt{1 +s} \left(1+\frac12 s\right)} \, ds$$

Maybe this symmetry could help.

Substituting $s \to -s$ we have:

$$J=\int_{-\frac{2}{3}}^0 \frac{\arctan \frac{1}{\sqrt{1-s} }}{\sqrt{1 +s} \left(1+\frac12 s\right)} \, ds$$

$$\arctan \frac{1}{\sqrt{1-s} }= \frac{\pi}{2}-\arctan \sqrt{1-s}$$

$$J= \frac{\pi}{2}\int_{-\frac{2}{3}}^0 \frac{ds}{\sqrt{1 +s} \left(1+\frac12 s\right)} -\int_{-\frac{2}{3}}^0 \frac{\arctan \sqrt{1-s}}{\sqrt{1 +s} \left(1+\frac12 s\right)} \, ds$$

$$J= \frac{\pi^2}{6} -\int_{-\frac{2}{3}}^0 \frac{\arctan \sqrt{1-s}}{\sqrt{1 +s} \left(1+\frac12 s\right)} \, ds$$

Adding the two expressions for $J$ we obtain:

$$2J=\frac{5\pi^2}{12}-\frac{2 \pi}{3} \arctan \sqrt{\frac{3}{5}} -\int_{-\frac{2}{3}}^{\frac{2}{3}} \frac{\arctan \sqrt{1-s}}{\sqrt{1 +s} \left(1+\frac12 s\right)} \, ds$$

$$J=\frac{5\pi^2}{24}-\frac{\pi}{3} \arctan \sqrt{\frac{3}{5}} -\int_{-\frac{2}{3}}^{\frac{2}{3}} \frac{\arctan \sqrt{1-s}}{\sqrt{1 +s} \left(2+s\right)} \, ds$$

Again, the symmetry might help with the last integral.

So now we need to show:

$$Y=\int_{-\frac{2}{3}}^{\frac{2}{3}} \frac{\arctan \sqrt{1-s}}{\sqrt{1 +s} \left(2+s\right)} \, ds=\frac{2\pi^2}{15}-\frac{\pi}{3} \arctan \sqrt{\frac{3}{5}}$$

Note that a related integral (from numerical results):

$$\int_{-1}^{1} \frac{\arctan \sqrt{1-s}}{\sqrt{1 +s} \left(2+s\right)} \, ds= \frac{\pi^2}{6}$$

---

Update:

Turning arctangent into an integral, substituting $s=\sin \theta= \frac{2t}{1+t^2}$ and then integrating a rational integral w.r.t. $t$ (with Mathematica's help), I have made yet another form of the conjecture:

Prove that: $$\int_0^1 \frac{\arctan \frac{2 \sqrt{1+2 p^2}}{\sqrt{5} (1+p^2)}}{\sqrt{1+2 p^2} (1+3 p^2)} dp= \frac{\pi}{2} \arctan \sqrt{\frac{3}{5}}- \frac{\pi^2}{15}$$

This one looks more complicated, but at least the limits are nice.

The integral is similar to $B$ from Fengshan Xiong's solution.