Proving $\int_{\sqrt{5/7}}^1 \frac{(\pi-3\arctan\sqrt{\frac{2x^2-1}{3x^2-2}})\arctan x}{\sqrt{2x^2-1}(3x^2-1)} dx = \frac{\pi^3}{672}$

Question

Half a year ago I posted a problem here, in which a remarkable result is proved by a unusual (and nontrivial) approach: $$\int_{\sqrt{\frac{3}{5}}}^1 \frac{\arctan (x)}{\sqrt{2 x^2-1} \left(3 x^2-1\right)} \, dx=\frac{3 \pi ^2}{160}$$ I collected it from a Ramanujan-like blog where produce only formulas but no proofs. If fact, the blogger states a harder version of this kind of 'pathological' integral: $$\pi \int_{\sqrt{5/7}}^1 \frac{\arctan y}{\left(3 y^2-1\right)\sqrt{2 y^2-1}} \, dy-3 \int_{\sqrt{5/7}}^1 \frac{\arctan (y) \arctan\sqrt{\frac{2 y^2-1}{3 y^2-2}}}{\left(3 y^2-1\right) \sqrt{2 y^2-1}} \, dy=\frac{\pi ^3}{672}$$ My question is: how to prove the second identity? Even armed with weapons offered in the link, I haven't found a possible way. Any kind of help will be appreciated.

Answer 1

Let $I$ denotes the integral, then $$I = \frac{1}{2}\int_{5/7}^1 {\frac{{\arctan \sqrt x }}{{\sqrt x (3x - 1)\sqrt {2x - 1} }}(\pi - 3\arctan \sqrt {\frac{{2x - 1}}{{3x - 2}}} )dx} $$ let $x = \frac{{3 + u}}{{5 + u}}$, then $\arctan\sqrt{{u^2} - 1} = \pi - 2\arctan \sqrt {\frac{{2x - 1}}{{3x - 2}}}$, so $$I = \frac{1}{4}\int_2^\infty {\frac{{\arctan \sqrt {\frac{{3 + u}}{{5 + u}}} }}{{\sqrt {1 + u} (2 + u)\sqrt {3 + u} }}(3\arctan \sqrt {{u^2} - 1} - \pi )du} $$ Note that, for $u>2$, $$\int_{u/2}^{u - 1} {\frac{{dv}}{{\sqrt {u - v} \sqrt v (1 + u - v)(1 + v)}}} = \frac{{3\arctan \sqrt {{u^2} - 1} - \pi }}{{\sqrt {1 + u} (2 + u)}}$$ so $$I = \frac{1}{4}\int_2^\infty {\int_{u/2}^{u - 1} {\frac{{\arctan \sqrt {\frac{{3 + u}}{{5 + u}}} }}{{\sqrt {3 + u} }}\frac{1}{{\sqrt {u - v} \sqrt v (1 + u - v)(1 + v)}}} dvdu} $$ change of variables $u=x+y, v=y$ gives $$\begin{aligned}I &= \frac{1}{4}\int_1^\infty {\int_x^\infty {\frac{{\arctan \sqrt {\frac{{3 + x + y}}{{5 + x + y}}} }}{{\sqrt {3 + x + y} }}\frac{1}{{\sqrt x \sqrt y (1 + x)(1 + y)}}} dxdy} \\ &= \frac{1}{2}\int_1^\infty {\int_1^\infty {\frac{{\arctan \sqrt {\frac{{3 + {x^2} + {y^2}}}{{5 + {x^2} + {y^2}}}} }}{{\sqrt {3 + {x^2} + {y^2}} }}\frac{1}{{(1 + {x^2})(1 + {y^2})}}} dxdy} \qquad \text{(By symmetry)} \\ &= \frac{1}{2}\int_1^\infty {\int_1^\infty {\int_0^1 {\frac{1}{{\sqrt {4 + {x^2} + {y^2} + {z^2}} }}} \frac{{dxdydz}}{{(1 + {x^2})(1 + {y^2})(1 + {z^2})}}} } = \frac{f(1,2)}{2} \end{aligned}$$ where for $n_1,n_2\geq 0$, $n=n_1+n_2$, $f(n_1,n_2)$ is the $n$-dimensional integral, $$f(n_1,n_2) = \int_{{{[0,\pi /4]}^{{n_1}}}{{\times [\pi /4,\pi /2]}^{n_2}}} {\frac{1}{{{{(1 + {{\sec }^2}{x_1} + ... + {{\sec }^2}{x_n})}^{1/2}}}}d{x_i}} $$

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I will show that $f(n_1,n_2)/\pi^n \in \mathbb{Q}$, and give a recurrence formula for it, from which $\color{red}{f(1,2) = \frac{\pi^3}{336}}$ is obtained, thereby completing the proof.

(Lemma) Let $n_1,n_2$ be nonnegative integers, $n=n_1+n_2$, $m,r>0$. If $mr=n+1$, then $$\int_{{{[0,1]}^{{n_1}}}{{\times[0,\infty ]}^{{n_2}}}} {\frac{1}{{{{(1 + {x_1}^r + ... + {x_n}^r)}^m}}}d{x_i}} = \frac{r}{{\Gamma (m)}}\frac{{\Gamma {{(1 + \frac{1}{r})}^{n + 1}}}}{{{n_1} + 1}} $$

Proof: Let $f(x) = \int_0^x e^{-t^r}dt$, then $$\begin{aligned} &\quad\int_{{{[0,1]}^{{n_1}}}{{\times [0,\infty ]}^{{n_2}}}} {\frac{1}{{{{(1 + {x_1}^r + ... + {x_n}^r)}^m}}}d{x_i}}\\ &= \frac{1}{{\Gamma (m)}}\int_0^\infty {\int_{{{[0,1]}^{{n_1}}}{{[0,\infty ]}^{{n_2}}}} {{t^{m - 1}}{e^{ - (1 + {x_1}^r + ... + {x_n}^r)t}}dt} } \\ &= \frac{1}{{\Gamma (m)}}\int_0^\infty {{t^{m - 1}}{e^{ - t}}{{\left( {\int_0^\infty {{e^{ - {x^r}t}}dx} } \right)}^{{n_2}}}{{\left( {\int_0^1 {{e^{ - {x^r}t}}dx} } \right)}^{{n_1}}}dt} \\ &= \frac{{f{{(\infty )}^{{n_2}}}}}{{\Gamma (m)}}\int_0^\infty {{t^{m - 1}}{t^{ - n/r}}{e^{ - t}}{{\left( {\int_0^{{t^{1/r}}} {{e^{ - {x^r}}}dx} } \right)}^{{n_1}}}dt} \\ &= \frac{{\Gamma {{(1 + \frac{1}{r})}^{{n_2}}}r}}{{\Gamma (m)}}\int_0^\infty {{t^{mr - n - 1}}{e^{ - {t^r}}}f{{(t)}^{{n_1}}}dt} \end{aligned}$$ if $mr=n+1$, then the antiderivative of integrand is $f(x)^{n_1+1}/(n_1+1)$, the result follows. QED

Now let $$\begin{aligned}S &= \{(x,y)\subset \mathbb{R}^2 | 0\leq x,y\leq 1\} \\ T &= \{(x,y)\subset \mathbb{R}^2 | 0\leq y\leq x\leq 1\} \\ R &= \{(x,y)\subset \mathbb{R}^2 | 0\leq x \leq 1, y\geq x\} \\ U &= \{(x,y)\subset \mathbb{R}^2 | 0\leq x \leq 1, y\geq 0\} \end{aligned}$$ note that under polar coordinates, $T$ and $R$ correspond to $0\leq r \leq \sec \theta, 0\leq \theta \leq \pi/4$ and $0\leq r \leq \sec \theta, \pi/4 \leq \theta \leq \pi/2$ respectively. For any (measurable) set $A$, let $$m(A) = \int_{A} \frac{dx_i}{(1+x_1^2+\cdots+x_{2n}^2)^{(2n+1)/2}} $$ this is symmetric under permutation of each $2n$ coordinates, consider ($n=n_1+n_2$) $$\begin{aligned}m({T^{{n_1}}} \times {R^{{n_2}}}) &= m({T^{{n_1}}} \times {(U - T)^{{n_2}}}) \\ & = \sum\limits_{k = 0}^{{n_2}} {\binom{n_2}{k}{{( - 1)}^k}m({T^{{n_1} + k}} \times {U^{{n_2} - k}})} = \sum\limits_{k = 0}^{{n_2}} {\binom{n_2}{k}\frac{{{{( - 1)}^k}}}{{{2^{{n_1} + k}}}}m({S^{{n_1} + k}} \times {U^{{n_2} - k}})} \\ & = \sum\limits_{k = 0}^{{n_2}} {\binom{n_2}{k}\frac{{{{( - 1)}^k}}}{{{2^{{n_1} + k}}}}m({{[0,1]}^{2{n_1} + {n_2} + k}} \times {{[0,\infty ]}^{{n_2} - k}})} \\ \end{aligned}$$ Lemma implies $$\tag{1}m({T^{{n_1}}} \times {R^{{n_2}}}) = \frac{{\Gamma {{(\frac{3}{2})}^{2n + 1}}}}{{\Gamma (\frac{{2n + 1}}{2})}}\sum\limits_{k = 0}^{{n_2}} {\binom{n_2}{k}\frac{{{{( - 1)}^k}}}{{{2^{{n_1} + k}}}}\frac{2}{{2{n_1} + {n_2} + k + 1}}}$$ On the other hand, polar coordinates give, by integrating all $r_i$, $$\begin{aligned}m({T^{{n_1}}} \times {R^{{n_2}}}) &= \int_{{{[0,\sec {\theta _i}]}^n} \times {{[0,\pi /4]}^{{n_1}}} \times {{[\pi/4,\pi /2]}^{{n_2}}}} {\frac{{{r_1}...{r_n}d{r_i}d{\theta _i}}}{{{{(1 + {r_1}^2 + ... + {r_n}^2)}^{(2n + 1)/2}}}}}\\ &=\frac{1}{{(2n - 1)(2n - 3)...(1)}} {\sum\limits_{i,j \ge 0} {{{(\frac{\pi }{4})}^{{n_1} + {n_2} - i - j}}{{( - 1)}^{i + j}}\binom{n_1}{i}\binom{n_2}{j}f(i,j)} } \end{aligned}$$ Compare this with $(1)$, denoting $f(i,j) = (\pi/4)^{i+j} \tilde{f}(i,j)$ gives $$\tag{2}{2^{n - 1}}\sum\limits_{k = 0}^{{n_2}} {\binom{n_2}{k}\frac{{{{( - 1)}^k}}}{{{2^{{n_1} + k}}}}\frac{2}{{2{n_1} + {n_2} + k + 1}}} = \sum\limits_{i,j \ge 0} {{{( - 1)}^{i + j}}\binom{n_1}{i}\binom{n_2}{j}\tilde{f}(i,j)} $$

This is our desired recurrence, starting with $f(0,0)=1, f(1,0)=\pi/6, f(0,1)=\pi/12$, we can computes $f(i,j)$ for $i+j=2$, for example, letting $n_1=1, n_2=1$ in $(2)$ gives $f(1,1)$. The following are values of $f(i,j)$ for $i+j\leq 3$: $$\begin{aligned}&f(0,0)=1 \\ &f(1,0)=\pi/6 \quad f(0,1)=\pi/12 \\ &f(2,0)=\pi^2/30\quad f(1,1)=3\pi^2/160\quad f(0,2)=\pi^2/80\\ &f(3,0)=\pi^2/140\quad f(2,1)=29\pi^3/6720\quad \color{red}{f(1,2)=\pi^3/336}\quad f(0,3)=\pi^3/448 \\ \end{aligned}$$

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Remarks:

• It can be shown that $f(n,0)=\frac{n!^2}{(2n+1)!}\pi^n$.

• Mathematica code to calculate $f(n_1,n_2)$:

int[n10_Integer, n20_Integer] /; Min[n10, n20] >= 0 := Block[{n1 = n10, n2 = n20, f, n, g, m, r}, n = n1 + n2; f[0, 0] = 1; g[a_, b_] := 2^(a + b - 1)* Sum[Binomial[b, k] (-1)^k/2^(a + k)*2/(2 a + b + k + 1), {k, 0, b}]; For[m = 1, m <= n, m++, For[r = 0, r <= m, r++, f[r, m - r] = (-1)^ m*(g[r, m - r] - Plus @@ ((-1)^(#1 + #2)*Binomial[r, #1]*Binomial[m - r, #2]* f[#1, #2] & @@@ Most@Flatten[Table[{i, j}, {i, 0, r}, {j, 0, m - r}], 1]))]]; f[n1, n2]*(Pi/4)^n];

For example, int[12,6] gives $f(12,6) = \frac{807986899 \pi ^{18}}{795635388421609881600}$.

• Consider $f(1,1)$, letting $y = \sqrt{(2+x^2)/(4+x^2)}$ gives $$f(1,1)=\frac{3\pi^2}{160}=\int_1^\infty {\frac{{\arctan \sqrt {\frac{{2 + {x^2}}}{{4 + {x^2}}}} }}{{(1 + {x^2})\sqrt {2 + {x^2}} }}dx} = \int_{\sqrt {3/5} }^1 {\frac{{\arctan y}}{{(3{y^2} - 1)\sqrt {2{y^2} - 1} }}dy} $$ This is a more direct proof of this non-trivial result, without using Schläfli's $S(\alpha,\beta,\gamma)$. The current question can be seen as a higher-dimensional analogue of such already difficult result.

Answer 2

I know I'm pretty late to this question, but I did figure out a method to simplify Pisco's recurrence relationship.

I will be using generating functions to solve the problem, though there might be better ways to finish this.

We will start by defining:

$$F(x,y)=\sum_{i,j\geq0}\tilde{f}(i,j)x^iy^j$$

Then we will multiply the recursive relationship by $x^{n_1}y^{n_2}$ and sum for all non-negative integer $n_1,n_2$:

$$\sum_{n_1,n_2\geq0}\bigg(\sum_{i,j\geq0}(-1)^{i+j}\binom{n_1}{i}\binom{n_2}{j}\tilde{f}(i,j)\bigg)x^{n_1}y^{n_2}=\\\sum_{n_1,n_2\geq0}\sum_{k=0}^{n_2}\binom{n_2}{k}(-1)^k2^{n_2-k}\frac{1}{2n_1+n_2+k+1}x^{n_1}y^{n_2}$$

I will now simplify the left-hand side first using binomial expansion, and the calculation will be separated out for ease of reading:

$$\sum_{n_1,n_2\geq0}\bigg(\sum_{i,j\geq0}(-1)^{i+j}\binom{n_1}{i}\binom{n_2}{j}\tilde{f}(i,j)\bigg)x^{n_1}y^{n_2}\\ =\sum_{i,j\geq0}(-1)^{i+j}\tilde{f}(i,j)\sum_{n_1=i}^\infty\binom{n_1}i{x^{n_1}}\sum_{n_2=j}^\infty\binom{n_2}j{y^{n_2}}\\ =\sum_{i,j\geq0}(-1)^{i+j}\tilde{f}(i,j)\frac{x^i}{(1-x)^{i+1}}\frac{y^j}{(1-y)^{j+1}}\\ \frac{1}{(1-x)(1-y)}F\bigg(\frac{x}{x-1},\frac{y}{y-1}\bigg)$$

We can now substitute back into our original equation, and perform a suitable substitution to get back $F(x,y)$ and once again using more binomial expansion:

$$\frac{1}{(1-x)(1-y)}F\bigg(\frac{x}{x-1},\frac{y}{y-1}\bigg)=\sum_{n_1,n_2\geq0}\sum_{k=0}^{n_2}\binom{n_2}{k}(-1)^k2^{n_2-k}\frac{1}{2n_1+n_2+k+1}x^{n_1}y^{n_2}\\ F(x,y)=\frac{1}{(1-x)(1-y)}\sum_{n_1,n_2\geq0}\sum_{k=0}^{n_2}\binom{n_2}{k}(-1)^k2^{n_2-k}\frac{1}{2n_1+n_2+k+1}\bigg(\frac{x}{x-1}\bigg)^{n_1}\bigg(\frac{y}{y-1}\bigg)^{n_2}\\ =\sum_{n_1,n_2\geq0}\sum_{k=0}^{n_2}\binom{n_2}{k}(-1)^{n+k}2^{n_2-k}\frac{1}{2n_1+n_2+k+1}x^{n_1}y^{n_2}\sum_{m_1=0}^\infty\binom{n_1+m_1}{n_1}x^{m_1}\sum_{m_2=0}^\infty\binom{n_2+m_2}{n_2}y^{m_2}\\ =\sum_{n_1,n_2\geq0}\sum_{m_1,m_2\geq0}\sum_{k=0}^{n_2}\binom{n_2}{k}\binom{n_1+m_1}{n_1}\binom{n_2+m_2}{n_2}(-1)^{n+k}2^{n_2-k}\frac{1}{2n_1+n_2+k+1}x^{n_1+m_1}y^{n_2+m_2}$$

This is now a massively complicated expression, but in order to make the power of $x,y$ depend on one variable each, we can perform the following substitution: $n_i \to m_i$ and $n_i+m_i \to n_i$. Thus, we can get the following:

$$F(x,y)=\sum_{n_1,n_2\geq0}\sum_{m_1=0}^{n_1}\sum_{m_2=0}^{n_2}\sum_{k=0}^{m_2}\binom{m_2}{k}\binom{n_1}{m_1}\binom{n_2}{m_2}(-1)^{m_1+m_2+k}2^{m_2-k}\frac{1}{2m_1+m_2+k+1}x^{n_1}y^{n_2}$$

We can now clearly see our original $\tilde{f}(n_1,n_2)$ as the inner 3 summations by the definition of $F(x,y)$.

$$\tilde{f}(n_1,n_2)=\sum_{m_1=0}^{n_1}\binom{n_1}{m_1}(-1)^{m_1}\sum_{m_2=0}^{n_2}\binom{n_2}{k}(-1)^{m_2}2^{m_2}\sum_{k=0}^{m_2}\binom{n_2}{m_2}(-1)^{k}2^{-k}\frac{1}{2m_1+m_2+k+1}$$

We will solve the inner most summation by way of changing the fraction into an integral, and again using more binomial expansion:

$$\sum_{k=0}^{m_2}\binom{n_2}{m_2}(-1)^{k}2^{-k}\frac{1}{2m_1+m_2+k+1}=\sum_{k=0}^{m_2}\binom{n_2}{m_2}(-1)^{k}2^{-k}\int_0^1{t^{2m_1+m_2+k}dt}\\ =\int_0^1t^{2m_1+m_2}\sum_{k=0}^{m_2}\binom{n_2}{m_2}(-1)^{k}2^{-k}{t^{k}dt}=\int_0^1t^{2m_1+m_2}\sum_{k=0}^{m_2}\binom{n_2}{m_2}{\bigg(\frac{-t}{2}\bigg)^{k}dt}\\ =\int_0^1t^{2m_1+m_2}\bigg(1-\frac{t}{2}\bigg)^{m_2}dt$$

Then, we can repeat the above process for the 2nd summation:

$$\sum_{m_2=0}^{n_2}\binom{n_2}{k}(-1)^{m_2}2^{m_2}\int_0^1t^{2m_1+m_2}\bigg(1-\frac{t}{2}\bigg)^{m_2}dt=\int_0^1t^{2m_1}\sum_{m_2=0}^{n_2}\binom{n_2}{k}(-2t)^{m_2}\bigg(1-\frac{t}{2}\bigg)^{m_2}dt\\ =\int_0^1t^{2m_1}\sum_{m_2=0}^{n_2}\binom{n_2}{k}(-2t-t^2)^{m_2}dt=\int_0^1t^{2m_1}(1-2t+t^2)^{m_2}dt$$

At this point, we can actually convert the formula into summation as we can recognise the integral is in fact the Beta Function:

$$\tilde{f}(n_1,n_2)=\sum_{m_1=0}^{n_1}\binom{n_1}{m_1}(-1)^{m_1}\int_0^1t^{2m_1}(1-2t+t^2)^{m_2}dt\\ =\sum_{m_1=0}^{n_1}\binom{n_1}{m_1}(-1)^{m_1}\int_0^1t^{2m_1}(1-t)^{2m_2}dt=\sum_{m_1=0}^{n_1}\binom{n_1}{m_1}(-1)^{m_1}B(2m_1+1,2m_2+1)=\sum_{m_1=0}^{n_1}\binom{n_1}{m_1}(-1)^{m_1}\frac{(2m_1)!(2n_2)!}{(2m_1+2n_2+1)!}$$

The above answer is already much better than the recursion formula needed to calculate our values, but we can get an alternative hypergeometric function solution that will not require any summation.

To be specific, we will be using the Euler type integral form of the hypergeometric function:

$$\int_0^1x^{b-1}(1-x)^{c-b-1}(1-zx)^{-a}dx=B(b,c-b)\ _2F_1(a,b;c;z)$$

Therefore we get:

$$\tilde{f}(n_1,n_2)=\int_0^1(1-t)^{2m_2}\sum_{m_1=0}^{n_1}\binom{n_1}{m_1}(-t^2)^{2m_1}dt\\ =\int_0^1(1-t)^{2m_2}(1-t^2)^{}dt=\int_0^1(1-t)^{2m_2+m_1}(1-t)^{m_1}dt\\ =\int_0^1u^{2m_2+m_1}(2-t)^{m_1}dt\\ =2^{n_1}B(n_1+2n_2+1,1)\ _2F_1\bigg(-n_1,n_1+2n_2+1;n_1+2n_2+2;\frac{1}{2}\bigg)\\ =\frac{2^{n_1}}{n_1+2n_2+1}\ _2F_1\bigg(-n_1,n_1+2n_2+1;n_1+2n_2+2;\frac{1}{2}\bigg)$$

One final trick we can do Pfaff transformation to simplify our function by a small amount:

$$\tilde{f}(n_1,n_2)=\frac{\ _2F_1(-n_1,1;n_1+2n_2+2;-1)}{n_1+2n_2+1}$$

Hence, we get our final answer for the general form of $f(n_1,n_2)$:

$$\bbox[5px,border:2px solid black] {f(n_1,n_2)=\bigg(\frac{\pi}{4}\bigg)^{n_1+n_2}\frac{\ _2F_1(-n_1,1;n_1+2n_2+2;-1)}{n_1+2n_2+1}}$$

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Remarks:

• The Mathematica code to calculate $f(n_1,n_2)$ using the above method

int4[n11_Integer,n12_Integer] /; Min[n11, n12] >=0 := (Pi/4)^(n11+n12)*Hypergeometric2F1[-n11,1,n11+2n12+2,-1]/(2n12+n11+1)