Let $\mathcal{N}(\mu,\sigma^2)$ denote the Gaussian distribution on $\mathbb{R}$:

$$ \mathcal{N}(\mu,\sigma^2)(x) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}.$$

A Gaussian distribution is defined by its mean $\mu\in\mathbb{R}$ and its standard deviation $\sigma>0$. Thus the upper-half plane is the parameter space for the normal family of distributions. Let $z = \mu+i\sigma$ denote an element of the upper-half plane, to be interpreted as the parameters for the Gaussian $\mathcal{N}(\mu,\sigma^2)$.

Let $G\in SL(2,\mathbb{R})$, the group of two by two real matrices with unit determinant. Let $g = \begin{pmatrix}a & b\\ c&d\end{pmatrix}$ be an element of $G$. This group acts on the upper-half plane by fractional linear transformations:

$$g\cdot z = \frac{az+b}{cz+d}.$$

Thus $G$ acts on Gaussians via $$ g\cdot \mathcal{N}(\mu,\sigma^2) = \mathcal{N}(Re(g\cdot z),\, (Im(g\cdot z))^2).$$

We now describe this action in more detail. If $g = \begin{pmatrix}a & b\\ 0&d\end{pmatrix}$, then $g$ acts on a Gaussian as follows: $$ g\cdot \mathcal{N}(\mu,\sigma^2) = \mathcal{N}\left(\frac a d \mu + \frac b d,\, \frac {a^2} {d^2} \sigma^2\right).$$

For an arbitrary $g$, it follows that $$\begin{align*} Re(g\cdot z) &= \frac{ac(\mu^2+\sigma^2) + bd + (ad+bc)\mu}{(c\mu+d)^2+c^2\sigma^2}\\ Im(g\cdot z) &= \frac{\sigma}{(c\mu+d)^2+c^2\sigma^2}, \end{align*} $$ and the action is $$ g\cdot \mathcal{N}(\mu,\sigma^2) = \mathcal{N}\left(\frac{ac(\mu^2+\sigma^2) + bd + (ad+bc)\mu}{(c\mu+d)^2+c^2\sigma^2},\left(\frac{\sigma}{(c\mu+d)^2+c^2\sigma^2}\right)^2\right).$$

$G$ has the Iwasawa decomposition $KAN$. Thus $g = k(g)a(g)n(g)$, where $$\begin{align*} k(g) &= \begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta&\cos\theta\end{pmatrix}\end{align*}\\ a(g) = \begin{pmatrix}e^t & 0\\ 0&e^{-t}\end{pmatrix}\\ n(g) = \begin{pmatrix}1 & u\\ 0&1\end{pmatrix}, $$ where $\theta$, $t$, and $u$ are analytic functions of the coefficients of $g$ (see Keith Conrad's notes).

It follows that $N$ acts on a Gaussian by moving its mean: $$ n(g)\cdot \mathcal{N}(\mu,\sigma^2) = \mathcal{N}\left( \mu + u\,, \sigma^2\right),$$ and $A$ acts on a Gaussian by dilating the mean and standard deviation: $$ a(g)\cdot \mathcal{N}(\mu,\sigma^2) = \mathcal{N}\left( e^{2t}\mu\,, e^{4t}\sigma^2\right).$$

$K$'s action is much more interesting:

$$ k(g)\cdot \mathcal{N}(\mu,\sigma^2) = \mathcal{N}\left(\frac{\cos\theta\sin\theta(\mu^2+\sigma^2) - \cos\theta\sin\theta + (\cos 2\theta)\mu}{((\sin\theta)\mu+\cos\theta)^2+(\sin^2\theta)\sigma^2}, \left(\frac{\sigma}{((\sin\theta)\mu+\cos\theta)^2+(\sin^2\theta)\sigma^2}\right)^2\right).$$

The standard normal $\mathcal{N}(0,1)$ is fixed by $K$. For any other Gaussian $\mathcal{N}(\mu,\sigma^2)$, moving through values of $\theta$ will cause oscillations of the mean and standard deviation, creating a "wobbling" of the Gaussian.

Questions: What are some physical interpretations of this? What are some good resources for exploring similar links among probability and geometry?

Let $X$ and $Y$ be two quandles and $f: X \rightarrow Y$ be a quandle homomorphism. Then we can define a map $\bar f: Inn(X) \rightarrow Inn(Y)$ as $\bar f(S_a)=S_{f(a)}$, where $a \in X$. Then $\bar f$ may not be a group homomorphism. But I am not able to construct such example.

I have posted this question on Math Stack Exchange, but have not got any reply.The link is here https://math.stackexchange.com/questions/2859327/quandle-homomorphism-does-not-always-induces-group-homomorphim-on-inner-automorp.

Any suggestions or hints would be of great help.

I want to compute $$\max_{\frac{1}{5 \theta }\leq \alpha \leq \frac{1}{2}} \left(\frac{\alpha\log (\alpha)}{1-\alpha} + \log \left( 1 - \alpha\right) + \frac{1}{1-\alpha} \cdot \left( f\left(\frac{1-\alpha}{2\alpha}\right) + 1 + \frac{41}{15\theta} \right)\right)$$ for \begin{align*} f(x) = \begin{cases} - 1.3- \log x + x/2 +1/2 - 1/x&\text{for } x \geq 2\\ -1.3 &\text{otherwise}. \end{cases} \end{align*}

Intuitively, it should be either for $\alpha=1/2$ or for $\alpha = 1/(5\theta)$ (and a plot verifies this). Is there a way how to prove this?

In the end I want to compute the $\theta$ where the whole expression reaches its minimum. Could there be another approach to do this?

Thanks in advance!!

$T_n$ be the full transformation semigroup on $X_n= \{1, 2, \cdots , n\}$.

$D_r =\{\alpha \in T_n: |im(\alpha)|=r\}$.

$E(D_r)$ is the set of all idempotents of semigroup $T_n$.

$support(\alpha)=\{x: x\alpha\neq x\}$

We know that $\alpha \in T_n$ is idempotent $\iff$ $x\alpha=x$ for all $x\in im(\alpha)$.

I wonder product of idempotents. For example; $\alpha =\begin{pmatrix} 1 & 2 & 3 & 4 \\ 1 & 2 & 2 & 1 \end{pmatrix}$, $\beta=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 3 & 4 \end{pmatrix} \in E(D_2)$, but $\alpha\beta=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 4 & 3 \end{pmatrix}$ is not idempotent.

$\textbf{Question:}$ Let $\alpha_0, \alpha_1\cdots, \alpha_k\in E(D_r)$ and $support(\alpha_i)\cap support(\alpha_j)=\emptyset, 0\leq i\neq j\leq k $.

Under which condition the product $\alpha_0\alpha_1\cdots \alpha_k$ becomes idempotent ?

Suppose $u(t,x)$ is a smooth velocity field on $[0,\infty)\times \mathbb{R}$ and periodic in space, i.e., $u(t,0)=u(t,1)$ $\forall t$. Assume that $\int_0^1 u(t,x) \,dx = c$, independent of time. Let $x(t)$ be the trajectory with $x(0)=0$. Prove that $\displaystyle \lim_{t\to\infty} \lvert x(t)\rvert/t \le \lvert c\rvert.$ (Alternatively, reduce to the circle and ask about the winding number of a trajectory.)

I'm not sure if this is true, but if it is, it would complete the last step in a theorem I'm trying to prove about the $\mu$-Hunter-Saxton equation. The reason I think it's true is the following elementary argument in the time-independent case: if $u(x)$ has a zero, then some fixed point is attracting and this limit will be zero. Otherwise we can assume $u(x)$ is always positive and compute that the period is $\displaystyle T = \int_0^1 dx/u(x)$. Then by Cauchy-Schwarz we must have $Tc \ge (\int_0^1 \,dx)^2 = 1$, so that $\frac{1}{T}\le c$. But this argument seems cumbersome and doesn't work in the nonautonomous case. Also this seems like a thing that should be well-known.

Let $A$ be a finite type algebra over $\mathbb{C}$. Does there exist a finite type $\mathbb{C}$-algebra $B$ and a nonzero divisor $b \in B$ such that $B/b \cong A$ and $B[1/b]$ is Cohen-Macaulay (or, even better, regular)?

EDIT: In the view of the very helpful counterexample of Jason Starr given in the comments below, let us assume in addition that $A$ is (S$_2$).

Given a large integer $N$, a small $\epsilon>0$ and integers $A,B,C,D,E,F,G,H$ and prime p where

$A,B,C,D\approx N^{\frac14+\epsilon}$ with $A,B,C,D$ mutually coprime

$E,F,G,H$ $\approx$ $N^{\alpha}$

$p$ is of size $N^{\frac12+\alpha+2\epsilon}$

hold what is the minimum $\alpha\geq\frac12$ such that for every pair $(m,m')\in\mathbb Z\times\mathbb Z$ with $p\nmid m$ and $p\nmid m'$ there are $a,b,c,d\in\mathbb Z$ such that $$m\equiv (ACa+BDb)(Gc+Hd)\bmod p$$ $$m'\equiv (ADc+BCd)(Ea+Fb)\bmod p$$ hold?

There are four parameters $a,b,c,d$ and only two targets $m,m'$ and four linear forms which are interleaved in the congruences. That is why I think it might be a pigeonhole type argument.

In principle there is sufficient room to hit $m$ and $m'$ in arithmetic progressions modulo $p$ however how to show this? Is this doable with exponential sums? At least can some $m,m'$ with $1\leq|m|,|m'|\leq N^\epsilon$ be representable? How small can $|mm'|$ be?

Essentially at large enough but fixed $\alpha\geq\frac12$ is there an $N_\alpha$ such that for every integer $N>N_\alpha$ and small enough $\epsilon$ if we choose $A,B,C,D,E,F,G,H,p$ in the way above then for every pair of $(m,m')\in\mathbb Z\times\mathbb Z$ we can find $a,b,c,d$ such that the congruence above works?

A Circular prime example is 193,939. Circular primes are numbers such that regardless of the what way you arrange the digits, the resulting number is a prime.

Consider the problem of minimising $$ J(u(.)) \triangleq \int_0^T l(x(t),u(t)) d t + \phi(x(T)), \: T> 0 $$ over a space of controls $\mathcal{U}$ with the constraint $$ \dot x(t) = f(x(t),u(t)), \: t >0, \: x(0) = x_0. $$ To make it simple, everything is $\mathbb{R}$ valued. We also assume that the functions $l,\phi,f$ and the space of controls $\mathcal{U}$ satisfy appropriate conditions for having a well-posed problem.

Under this form we can employ either Bellman or Pontryagin principles to study the optimal control.

**Question.** Can we / How to use this framework to impose a terminal value for $x$? say $x(T) = x_\star$.

Can shapes determined by number of points?

From an amazing theorem in plane curves geometry we know that vertices of triangles similar to arbitrary triangles $T$ is **dense** on every closed jordan curve in a plane, so if $J$ be such curve and $A,B,C$ be 3 noncollinear points on plane then at least one curve similar to $J$ contains $A,B,C$. If $J$ be a circle then exactly one circle passes through $A,B,C$. From here we reach the problem in its simplest form:

**QUESTION:** is circle the only shape on a **Euclidean** plane (not only from closed curves mentioned above) which **just one similar to it** defined by each set of 3 non-collinear points we consider from $\mathbb{R}^2$ ? *(Means not two or more similar of the shape fits $A,B,C$, just one of it).*

**Note:** here the mentioned "shape" can be any subset of $\mathbb{R}^2$. The "trivial triangle" trivially could not be an answer because we can find many similar of it passing through the 3 points creates it as vertices. It seems for an specified set of 3 points with predefined angles of the triangle it creates, also we may find just circle which is unique.

What about generalization to $n$ points in **$\mathbb{R}^m$** which exactly define **$k$** similar shapes?

As a follow-up to Automorphisms of the L-function associated to an elliptic $\mathbb{Q}$-curve, I would like to know if there is a natural way to define morphisms $ \phi $ of (isogeny classes of) elliptic curves such that :

There exists an isogeny $ f $ from $ E_1 $ to $ E_2 $ iff there exists an isogeny $ f_{\phi} $ from $ \phi(E_{1}) $ to $ \phi(E_{2}) $ that is equivalent to $ f $.

$ f_{\phi}\circ\phi=\phi\circ f $.

Building from Chris Wuthrich's comment below, it appears that the preservation of an isogeny class makes $ \phi $ an endofunctor of the considered category. As an isogeny class determines the L-function associated to a representant thereof, this endofunctor induces an automorphism of this L-function. The question is thus : does every automorphism of such an L-function arise this way ?

I want to determine the subset of $m$ members ($m < n/2$) of the set $e^{i 2\pi k/n}, \ \ k=0,\dots, n-1$, so that the absolute value of its sum is maximal.

This question is a follow up.

Let $A$ be a real square matrix of size $n \times n$. How to determine the minimum spectral norm under diagonal similarity, i.e.,

$$ s(A) = \min_{D} \lVert D^{-1} A D\rVert_2, $$ where $D$ is a non-singular, diagonal real matrix. As it is unlikely to find an analytical upper bound, I would like to ask how $s(A)$ could be determined numericallly.

I asked a question here order of a permutation and lexicographic order but it seems*** that a powerful and rich generalization can be made!

Let $A$ be a ring together with an arbitrary **good order** $<^*$ and let $L: M_n(A)\to M_n(A)$ be the application that sorts the rows of a matrix according to the lexicographic order (induced by $<^*$) and the columns of the matrice we get, according to the lexicographic order. We then define $L_Q(M):=L(M).Q$ for any $Q\in M_n(A)$.

[edit : I previously just asked $<^*$ to be a total order (see third comment )]

Suppose that $Q^q=Id$ for some $q\in \mathbb N$ and $Q\in GL_n(A)$. Is is true that for any $M\in M_n(A)$, there exists $r\in \mathbb N$ such that for any $i\in \mathbb N$, we have $L_Q^r(M)=L_Q^{r+iq}(M)$ ?

(***the experimentations I made with a computer concern $A=\mathbb F_2$)

[***edit : I also examined the case $A=\mathbb Z$ with $<^*$ defined by

$a<^*b$ iff ($2a^2+a/|a|<2b^2+b/|b|$) the usual order fail in some cases,

where $k\mapsto L_Q^k(M)$ is not almost-periodic, but the usual order it is NOT a good order, and the cases that failed with it are not failing anymore with $<^*$ as defined in this edit !]

Fix throughout a finite set $T \subset S^{d-1}$ intersecting every open hemisphere of $S^{d-1}$. A $T$-polytope is a (necessarily bounded) set $P$ of the form $$ P = \bigcap_{u \in T} \big\{ x \in \mathbb{R}^d : \langle x,u \rangle \leq a_u \big\}. $$ Given a bounded set $X \subset \mathbb{R}^d$, let $\mathrm{cont}(X)$ denote the minimal $T$-polytope containing $X$ (I'm thinking of this as the `container' of $X$).

Now let $A$ and $B$ be two $T$-polytopes such that $\mathbf{0} \in A \cap B$. One can show quite easily that in two dimensions we always have $\mathrm{cont}(A \cup B) \subset A + B$. On the other hand, one can construct examples in three or more dimensions where this relation does not hold. This (hopefully!) motivates the following slightly stranger question. Given a $T$-polytope $P$ as above, let us say that $x$ is a special point of $P$ if there are $d$ linearly independent vectors $u_1,\dots,u_d \in T$ such that $\langle x,u_i \rangle = a_{u_i}$ for all $i$. (In other words, the special points are the intersections of any d extended faces of $P$ in linearly independent directions.) Let $E(P)$ denote the convex hull of the set of all special points of $P$. Thus, as an example, if $P$ is a simplex or a parallelepiped, then $E(P) = P$. Clearly in general, $P \subset E(P)$, since the vertices of $P$ are special. My question is: with $A$ and $B$ as above, is it true that $E(\mathrm{cont}(A \cup B)) \subset E(A) + E(B)$?

Very roughly speaking, the function $E$ is an attempt to circumvent one of the reasons that the relation $\mathrm{cont}(A \cup B) \subset A + B$ fails in three (say) dimensions, which is that $A + B$ need not be a $T$-polytope (by contrast, in two dimensions, $A + B$ is always a $T$-polytope). By this I don't mean that $E(A) + E(B)$ is a $T$-polytope (it clearly isn't), but that the points in $E(A)$ (say) that are not vertices of $A$ seem to be the points that `hurt' us when we try to prove the false relation.

Final question: If the proposed containment is false, could there be another simple function of $T$-polytopes that does obey such a containment relation?

An algebraic curve defined by a polynomial of degree n can have at most (n−1)(n−2)/2 singularities. What is the maximal number of singularities of a curve in a k-dimensional space in terms of the degree of the k-1 polynomial equations defining the curve?

P.S.: The comment of Mohan posted below gives an upper bound on the number of singularities by using the fact the the geometric genus is non-negative. So, I could reformulate my question as follows. Are there space curves of genus 0 (or 1) such that the upper bound (possibly minus 1) is attained? given for granted that the curve is irreducible.

**Conjecture:** Let $f:\mathbb{R}→\mathbb{R}$ be an **everywhere** differentiable function and assume that $f(x)+f′(x)∈ \{-1,1\}$ **almost everywhere** and $f'(0)=0$. Then is $f$ necessarily a constant function?

Can you give me a counter-example? I have already asked the question here on MathSE.

I'm looking for suggestions on how one might try to compute the following $(N-1)$-dimensional integral:

$$I_N= \frac{1}{(2\pi)^{N-1}(N-1)!} \int\cdots\int \\ \begin{vmatrix} 1 & 1 & \cdots & 1 & 1 \\ e^{i\alpha_1} & e^{i\alpha_2} & \cdots & e^{i\alpha_{N-1}} & e^{-i(\alpha_1+\dots+\alpha_{N-1})} \\ e^{i2\alpha_1} & e^{i2\alpha_2} & \cdots & e^{i2\alpha_{N-1}} & e^{-i2(\alpha_1+\dots+\alpha_{N-1})} \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ e^{i(N-1)\alpha_1} & e^{i(N-1)\alpha_2} & \cdots & e^{i(N-1)\alpha_{N-1}} & e^{-i(N-1)(\alpha_1+\dots+\alpha_{N-1})} \\ \end{vmatrix}^2 {d\alpha_1\cdots d\alpha_{N-1}}$$

over the region given by the following conditions:

$$\alpha_{N-1}\in\left(-\pi, -\pi+\frac{2\pi}{N}\right) \text{ and } \alpha_1,\dots,\alpha_{N-2},-(\alpha_1+\dots+\alpha_{N-1})\in\left(-\pi+\frac{2\pi}{N}, \pi\right)$$

More generally, the integration region can be described as: one of the variables $\alpha_i, 1\leq i\leq N-1$ must be inside the interval $\left(-\pi, -\pi+\frac{2\pi}{N}\right)$, and the rest of the $\alpha_i$, together with $-\alpha_1-\dots-\alpha_{N-1}$ must be outside this interval $\mod 2\pi$. $\alpha_{N-1}$ is just a particular choice and the value of $I_N$ doesn't change if $\alpha_{N-1}$ is replaced by some other $\alpha_i$; it also doesn't change if the interval $\left(-\pi, -\pi+\frac{2\pi}{N}\right)$ gets replaced by $\left(\pi-\frac{2\pi}{N}, \pi\right)$

Regarding the shape of the integration region, in the $N=3$ case it is a hexagon, in the $N=4$ case it's an octahedron, and in general, the region for $I_N$ is an $(N-1)$-dimensional polytope. The normalization factor in front of the integral is just to ensure that $0\leq I_N\leq 1$

The integrand is the square of the absolute value of the determinant, and can also be expressed as $$\prod_{1\leq j<k\leq N-1} |e^{i\alpha_j}-e^{i\alpha_k}|^2 \prod_{1\leq j\leq N-1} |e^{i\alpha_j}-e^{-i(\alpha_1+\dots+\alpha_{N-1})}|^2$$

or equivalently

$$2^{N(N-1)} \prod_{1\leq j<k\leq N-1} \sin^2\frac{\alpha_j-\alpha_k}{2} \prod_{1\leq j\leq N-1} \sin^2\frac{\alpha_j+\alpha_1+\dots+\alpha_{N-1}}{2} $$

So far, I have computed the values for when $N=2,3,4$

\begin{align} I_2 & =1 \\[8pt] I_3 & =\frac{7}{12}+\frac{11}{3\pi^2} \\[8pt] I_4 & =\frac{5}{12}+\frac{1001}{360\pi^2}+\frac{4232}{675\pi^3} \end{align}

but I haven't been able to figure out a method for computing the integral in the general case.

The limits of integration for each variable $\alpha_i$ can be expressed explicitly if we split the integration region into two equal parts, say $R_1$ and $R_2$, for which we have the following limits

\begin{array}{l | l} R_1 & R_2 \\ \hline \alpha_{4k+1}\in \left( -\pi+\frac{2\pi}{N} , \pi-\frac{2\pi}{N}-\sum_{4k+2}^{N-1}\alpha_i \right) & \alpha_{4k+1}\in \left( -\pi-2k\frac{2\pi}{N}-\sum_{4k+2}^{N-1}\alpha_i, \pi \right) \\ \alpha_{4k+2}\in \left( -(2k+1)\frac{2\pi}{N}-\sum_{4k+3}^{N-1}\alpha_i , \pi \right) & \alpha_{4k+2}\in \left( -\pi+\frac{2\pi}{N}, -(2k+1)\frac{2\pi}{N} -\sum_{4k+3}^{N-1}\alpha_i \right) \\ \alpha_{4k+3}\in \left( -\pi-(2k+1)\frac{2\pi}{N}-\sum_{4k+4}^{N-1}\alpha_i , \pi \right) & \alpha_{4k+3}\in \left( -\pi+\frac{2\pi}{N} , \pi-(2k+2)\frac{2\pi}{N} -\sum_{4k+4}^{N-1}\alpha_i \right) \\ \alpha_{4k+4}\in \left( -\pi+\frac{2\pi}{N} , -(2k+2)\frac{2\pi}{N}-\sum_{4k+5}^{N-1}\alpha_i \right) & \alpha_{4k+4}\in \left( -(2k+2)\frac{2\pi}{N} -\sum_{4k+5}^{N-1}\alpha_i , \pi \right) \\ \end{array}

for $4k+1, 4k+2, 4k+3, 4k+4 \in \{1,2,\dots,N-2\}$, and for $N-1$:

\begin{array}{l | l | l} & R_1 & R_2 \\ \hline \text{if } N=2n & \alpha_{N-1}\in \left( -\pi, -\pi+\frac{2\pi}{N} \right) & \alpha_{N-1}\in \left( -\pi, -\pi+\frac{2\pi}{N} \right) \\ \text{if } N=4n+1 & \alpha_{N-1}\in \left( -\pi, -\pi+\frac{\pi}{N} \right) & \alpha_{N-1}\in \left( -\pi+\frac{\pi}{N},-\pi+\frac{2\pi}{N} \right) \\ \text{ if} N=4n+3 & \alpha_{N-1}\in \left( -\pi+\frac{\pi}{N},-\pi+\frac{2\pi}{N} \right) & \alpha_{N-1}\in \left( -\pi, -\pi+\frac{\pi}{N} \right) \\ \end{array}

With these conventions, the integral over $R_1$ is equal to the integral over $R_2$ and equal to one-half of the integral over the total region.

For example, when $N=3$

$\frac{1}{(2\pi)^22!}\int\int_{R_1}|\dots|^2d\alpha_1 d\alpha_2 = \frac{1}{(2\pi)^22!}\int_{-\pi+\frac{\pi}{3}}^{-\pi+\frac{2\pi}{3}}\int_{-\pi+\frac{2\pi}{3}}^{\pi-\frac{2\pi}{3}-\alpha_2} |\dots|^2d\alpha_1 d\alpha_2 = \frac{1}{2}I_3 $

$\frac{1}{(2\pi)^22!}\int\int_{R_2}|\dots|^2d\alpha_1 d\alpha_2 = \frac{1}{(2\pi)^22!}\int_{-\pi}^{-\pi+\frac{\pi}{3}}\int_{-\pi-\alpha_2}^{\pi} |\dots|^2d\alpha_1 d\alpha_2 = \frac{1}{2}I_3 $

and if $N=4$

$\frac{1}{(2\pi)^33!}\int\int\int_{R_1}|\dots|^2d\alpha_1 d\alpha_2 d\alpha_3 = \frac{1}{(2\pi)^33!}\int_{-\pi}^{-\pi+\frac{2\pi}{4}}\int_{-\frac{2\pi}{4}-\alpha_3}^{\pi}\int_{-\pi+\frac{2\pi}{4}}^{\pi-\frac{2\pi}{4}-\alpha_2-\alpha_3} |\dots|^2d\alpha_1 d\alpha_2 d\alpha_3 = \frac{1}{2}I_4 $

$\frac{1}{(2\pi)^33!}\int\int\int_{R_2}|\dots|^2d\alpha_1 d\alpha_2 d\alpha_3 = \frac{1}{(2\pi)^33!}\int_{-\pi}^{-\pi+\frac{2\pi}{4}}\int_{-\pi+\frac{2\pi}{4}}^{-\frac{2\pi}{4}-\alpha_3}\int_{-\pi-\alpha_2-\alpha_3}^{\pi} |\dots|^2d\alpha_1 d\alpha_2 d\alpha_3 = \frac{1}{2}I_4 $

This is how I computed the values for $I_3$ and $I_4$, but for $N\geq 5$ it gets too messy, and I don't think this would be the best approach for the general case.

Consider the unbounded integral operator $T:D(T) \subset L^2((0,\infty),x \ dx ) \rightarrow L^2((0,\infty),x \ dx)$ on its maximal domain $D(T)=\left\{f \in L^2((0,\infty),x \ dx); Tf \in L^2((0,\infty),x \ dx) \right\}$ such that for some $x_0 \in (0,\infty)$ we have

$$(Tf)(y) = \frac{1}{y-x_0} \int_{0}^{\infty} f(x) \frac{xy}{x^2+y^2+1} \ dx.$$

I would like to show that $1$ is not(!) in the point spectrum of this operator (this is what I observed numerically), i.e.

$(Tf)(x)=f(x)$ does not have a solution.

Observations:

One problem (and perhaps the decisive problem) seems to be the singularity at $y=x_0.$

If such a eigenfunction existed, then

$$f(y) = \frac{1}{y-x_0} \int_{0}^{\infty} f(x) \frac{xy}{x^2+y^2+1} \ dx.$$

For $f$ to be integrable at $y=x_0,$ the integral $\int_{0}^{\infty} f(x) \frac{x}{x^2+y^2+1} \ dx$ needs to have a zero at $y=x_0$ and $f$ changes sign.

I tried to construct a contradiction from this but failed so far.

Suppose that $1/2+it$ is not a zero of the Riemann zeta function $\zeta$, where $t \in \mathbb{R}$. Can $1/\zeta(1/2+it)$ be expressed as a Dirichlet series ?