12
$\begingroup$

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.

$\endgroup$
2
  • 2
    $\begingroup$ Your description made me curious which blog it is $\endgroup$
    – user632577
    Feb 28, 2020 at 8:29
  • 4
    $\begingroup$ The integral might look better like this:$$\int_\sqrt{\frac35}^1\frac{\pi\cot^{-1}\left({\sqrt{2-x^2}}\right)-3\cot^{-1}\left({\sqrt{2-x^2}}\right)\cot^{-1}\left({\sqrt{2-1/x^2}}\right)}{1+x^2}dx$$ $\endgroup$
    – Zacky
    Feb 28, 2020 at 10:23

1 Answer 1

19
+50
$\begingroup$

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, $$f(n_1,n_2) = \int_{{{[0,\pi /4]}^{{n_1}}}{{\times [\pi /4,\pi /2]}^{n_2}}} {\frac{1}{{{{(1 + {{\sec }^2}{x_1} + ... + {{\sec }^2}{x_n})}^{1/2}}}}d{x_i}} $$


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 $i+j\leq 3$: $$\begin{aligned}&f(0,0)=1 \\ &f(1,0)=\pi/6 \quad f(0,1)=\pi/12 \\ &f(2,0)=\pi^2/30\quad f(1,1)=3\pi^2/160\quad f(0,2)=\pi^2/80\\ &f(3,0)=\pi^2/140\quad f(2,1)=29\pi^3/6720\quad \color{red}{f(1,2)=\pi^3/336}\quad f(0,3)=\pi^3/448 \\ \end{aligned}$$


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.
$\endgroup$
2
  • 2
    $\begingroup$ Amazing work! $ \, $ $\endgroup$
    – NoName
    Mar 9, 2020 at 2:45
  • $\begingroup$ @NoName Thank you for appreciation. :) $\endgroup$
    – pisco
    Mar 9, 2020 at 4:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.