High dimensional generalizations of $\int_{\sqrt{2} }^{\sqrt{3} } \frac{\arctan(y)}{(y^2-1)\sqrt{y^2-2} } dy =\frac{5\pi^2}{96}$

Question

Let $Q=n_1+n_2+n_3+1$,$\mathbf{s}=(n_1,n_2,n_3)$. Define $$ A(\mathbf{s}) =\int_{D}\prod_{n=1}^{Q-1}\frac{1}{1+x_n^2}\int_{1}^{\infty}\left(Q+\sum_{n=1}^{Q-1}x_n^2 +y^2\right)^{-1}\mathrm{d}y \text{d}x_i. $$ Where $D=[0,\infty]^{n_1}\times[0,1]^{n_2} \times[1,\infty]^{n_3}\subset\mathbb{R}^{Q-1}$, $\mathrm{d}x_i =\prod_{n=1}^{Q-1} \text{d}x_n$.

• Special case. For $\mathbf{s}=(0,0,n_3)$, we have $$ A(\mathbf{s}) =\frac{1}{Q}\left ( \frac{\pi}{4} \right )^Q. $$
• Question 1 Prove $$ \pi^{-Q}A(\mathbf{s})\in\mathbb{Q}. $$

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My ultimate goal is to evaluate $A(\mathbf{s})$. Numerical calculations suggest that $$A(1,0,0)=\frac{\pi^2}{12} \quad A(0,1,0)=\frac{5\pi^2}{96}\quad A(0,0,1)=\frac{\pi^2}{32}$$

$$A(2,0,0)=\frac{\pi^3}{32} \quad A(1,1,0)=\frac{\pi^3}{80}$$

$$A(0,3,0)=\frac{93\pi^4}{35840} \qquad A(0,4,0)=\frac{193\pi^5}{322560}$$

Actually, if we explicit calculate the multiple integrals, it yields $$ \begin{aligned} &\int_{\sqrt{2} }^{\sqrt{3} } \frac{\arctan(y)}{(y^2-1)\sqrt{y^2-2} } \text{d}y =\frac{5\pi^2}{96},\\ &\frac{\pi}{6} \int_{\sqrt{3} }^{\sqrt{5} } \frac{\arctan(y)}{(y^2-1)\sqrt{y^2-2} } \text{d}y -\int_{2}^{\sqrt{5} } \frac{\displaystyle{\arctan (y)\arctan \sqrt{\frac{y^2-4}{y^2-2} } } }{(y^2-1)\sqrt{y^2-2} } \text{d}y=\frac{11\pi^3}{5760},\\ &\frac{\pi}{6} \int_{\sqrt{3} }^{\sqrt{5} } \frac{\arctan\left(y\sqrt{2+y^2} \right)}{(y^2-1)\sqrt{y^2-2} } \text{d}y -\int_{2}^{\sqrt{5} } \frac{\displaystyle{\arctan\left(y\sqrt{2+y^2}\right)\arctan \sqrt{\frac{y^2-4}{y^2-2} } } }{(y^2-1)\sqrt{y^2-2} } \text{d}y=\frac{\pi^3}{420}. \end{aligned} $$ Are there any other simple results? For $Q=4$, we may meet some 'troubles', such as this one: $$ \int_{0}^{1} \int_{1}^{\sqrt{2} } \frac{u\left(\pi-2\arctan\sqrt{u^4-1}-2\arctan \sqrt{\frac{u^2-1}{u^2+1} } \right) \arctan \sqrt{4+u^2+v^2} } {(1+v^2)\sqrt{1+u^2}(2+u^2) \sqrt{4+u^2+v^2} } \text{d}u\text{d}v. $$ I can hardly convert into a 'simple' form.

• Question 2. Can we evaluate a more general family of this kind of integrals? $$A(\alpha,\mathbf{s}) =\int_{D}\prod_{n=1}^{Q-1}\frac{1}{\alpha^2+x_n^2}\int_{1}^{\infty}\left(\alpha^2 Q+\sum_{n=1}^{Q-1}x_n^2 +y^2\right)^{-1}\mathrm{d}y \text{d}x_i.$$

Answer 1

I've found a direct way.
Let me give an example, $$ I=\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{1}{(1+x^2)(1+y^2)(1+z^2)\sqrt{4+x^2+y^2+z^2} }\text{d}x\text{d}y\text{d}z. $$ Use the fact that $$ \frac{1}{\sqrt{x} } =\frac{2}{\sqrt{\pi}}\int_{0}^{\infty}e^{-xt^2}\text{d}t $$, and rewrite the integral into $$I = \frac{2}{\sqrt{\pi}} \int_{0}^{\infty}e^{-4t^2} \left ( \int_{0}^{1} \frac{e^{-x^2t^2}}{1+x^2} \text{d}x \right )^3 \text{d}t.$$ The inner integral is simply equal to $$\int_{0}^{1} \frac{e^{-x^2t^2}}{1+x^2} \text{d}x =\frac{\pi}{4}e^{t^2}\left(1-\operatorname{erf}(t)^2\right).$$ Then,$$I=\frac{\pi^3}{32}\cdot\frac{1}{\sqrt{\pi}} \int_{0}^{\infty}e^{-t^2}\left(1-\operatorname{erf}(t)^2\right)^3\text{d}t.$$ Note that the integrand has a polynomial of $\operatorname{erf}(t)$. That is to say, if we set $$P(x)=a_n x^n+a_{n-1} x^{n-1}+...+a_{1} x+a_0$$, we get $$\begin{aligned} I & = \frac{\pi^3}{32}\cdot\frac{1}{\sqrt{\pi} } \int_{0}^{\infty}e^{-t^2}P(\operatorname{erf}(t))\text{d}t \\ & = \frac{\pi^3}{32}\cdot\frac{1}{\sqrt{\pi} } \int_{0}^{\infty}e^{-t^2}\sum_{k=0}^{n}a_k\operatorname{erf}(t)^k\text{d}t\\ &= \frac{\pi^3}{32}\cdot\frac{1}{\sqrt{\pi} } \sum_{k=0}^{n}a_k \int_{0}^{\infty}e^{-t^2}\operatorname{erf}(t)^k\text{d}t\\ &=\frac{\pi^3}{32}\cdot\frac{1}{\sqrt{\pi} } \sum_{k=0}^{n}a_k\cdot\frac{\sqrt{\pi} }{2(k+1)} \\ &=\frac{\pi^3}{64} \sum_{k=0}^{n} \frac{a_k}{k+1} \end{aligned}$$ The finite sum is easy to evaluate, and we conclude that $$I=\frac{\pi^3}{140} .$$

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These four integrals have the similar expression: $$\begin{aligned} &\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{1}{(1+x^2+y^2+z^2)^2}\text{d}x\text{d}y\text{d}z\\ &\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{1}{(1+x^2)(2+x^2+y^2+z^2)^{3/2}}\text{d}x\text{d}y\text{d}z\\ &\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{1}{(1+x^2)(1+y^2)(3+x^2+y^2+z^2)}\text{d}x\text{d}y\text{d}z\\ &\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{1}{(1+x^2)(1+y^2)(1+z^2)\sqrt{4+x^2+y^2+z^2}}\text{d}x\text{d}y\text{d}z. \end{aligned}$$ Their limits can rewrite as $[1,\infty],[0,\infty]$ and still have similar results, of course.

Answer 2

The claim is non-trivial. Let $$g(n_1,n_2) = \int_1^\infty\int_{{{[0,\pi /4]}^{{n_1}}}{{\times [\pi /4,\pi /2]}^{n_2}}} {\frac{d{x_i}}{{{1+a^2 + {{\sec }^2}{x_1} + ... + {{\sec }^2}{x_n}}}}} da$$ We need to prove $$\tag{*}g(n_1,n_2)\in \pi^{1+n_1+n_2} \mathbb{Q}$$

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The solution is largely in parallel to the one I wrote here. One should read that answer first. For any (measurable) set $A\subset \mathbb{R}^{2n}$, let $$\nu(A) = \int_1^\infty \int_{A} \frac{dx_i}{(1+a^2+x_1^2+\cdots+x_{2n}^2)^{n+1}} da$$ With $T,R$ as in the linked answer, mutatis mutandis, we have (with $n=n_1+n_2$), $$\tag{1}\nu({T^{{n_1}}} \times {R^{{n_2}}}) = \sum\limits_{k = 0}^{{n_2}} {\binom{n_2}{k}\frac{{{{( - 1)}^k}}}{{{2^{{n_1} + k}}}}\nu({{[0,1]}^{2{n_1} + {n_2} + k}} \times {{[0,\infty ]}^{{n_2} - k}})} $$

The following lemma proved there

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

implies $$\tag{2}\nu({{[0,1]}^{2{n_1} + {n_2} + k}} \times {{[0,\infty ]}^{{n_2} - k}}) = \frac{\pi^{n+1}}{2^{2n+1}n!} \frac{1}{(2n_1+n_2+k+2)(2n_1+n_2+k+1)}$$

On the other hand, polar coordinates on each pairs $(x_1,x_2), (x_3,x_4),\cdots$ gives, $$\begin{aligned}\nu({T^{{n_1}}} \times {R^{{n_2}}}) &= \int_1^\infty\int_{{{[0,\sec {\theta _i}]}^n} \times {{[0,\pi /4]}^{{n_1}}} \times {{[\pi/4,\pi /2]}^{{n_2}}}} {\frac{{{r_1}...{r_n}d{r_i}d{\theta _i}}}{{{{(1 + {r_1}^2 + ... + {r_n}^2)}^{n+1}}}}} da\\ &=\frac{1}{2^nn!} {\sum\limits_{i,j \ge 0} {{{(\frac{\pi }{4})}^{{n_1} + {n_2} - i - j}}{{( - 1)}^{i + j}}\binom{n_1}{i}\binom{n_2}{j}g(i,j)} } \end{aligned}$$

Compare this with $(1),(2)$ will give a recurrence of $g(i,j)$, which allows one to calculate all $g(i,j)$ starting from inital value $g(0,0) = \frac{\pi}{4}$, proving $(*)$.