# Tag Info

### Intuitive explanation why "shadow operator" $\frac D{e^D-1}$ connects logarithms with trigonometric functions?

This is basically a lightly transformed version of Euler's cotangent partial fraction expansion $$\pi \cot(\pi z) = \frac{1}{z} + \sum_{n=1}^\infty \frac{1}{z-n} + \frac{1}{z+n}$$ (the log derivative ...

### Do distance functionals separate probability measures?

No. Suppose that $\Omega$ consists of four points arranged in a square, where adjacent points have distance 1 between them and opposite points have distance 2. Specifically, if the points are labeled ...
Accepted

### Representation of the Dirac delta function

As usual in such examples, there is no need to integrate against a test function. One can simply use the fact that if a sequence (or net) of distributions converges in the distributional sense, then ...

### Intuitive explanation why "shadow operator" $\frac D{e^D-1}$ connects logarithms with trigonometric functions?

The op $$T_x = \frac{D_x}{e^{D_x}-1} = e^{b.D_x},$$ where $(b.)^n = b_n$ are the Bernoulli numbers, is (mod signs) often referred to as the Todd operator (maybe originally given that name by ...
Accepted

### On an asymptotic integral decay

I'll change $[-1,0]$ to $[0,1]$ (so $t\to -t$), which seems easier on the brain. $f(t)=e^{-1/t}$ is a counterexample: For $0<t<1/\sqrt{\lambda}$, we have $f\le e^{-\sqrt{\lambda}}$, so this part ...

Accepted

### Physical interpretation of the mellin transform variable?

Physical interpretation: To develop a physical intuition, this article might be informative: The power spectrum of the Mellin transformation with applications to scaling of physical quantities The ...
Accepted

Accepted

### A solution satisfying an integral inequality is bounded

Apparently, the question should be interpreted as follows: Let $y\colon[0,\infty)\to[0,\infty)$ be a measurable function satisfying the inequality \begin{equation} y(t)+\int_0^t y(s)\,ds\le c_1\int_0^...
Accepted

### Integral of a function changing sign

In fact, this conjecture fails to hold for $x$ in a right neighborhood of $0$. Indeed, let $g(x,s)$ denote the integrand. Note that $g(x,s)$ is continuous in $x\in[0,1)$ for each real $s>0$, and ...

### A density claim

This is not an answer to the question, but a proof of something much weaker: there is a sequence $c_m\to0$ such that if $\lVert g_m\rVert\leq c_m$ for all $m$, then the result from the question is ...
Following Jochen Wengenroth suggestion, I sketch the proof of the result from a comment. The claim follows from the following Lemma. If $\mu$ is a measure supported on $[0,X]$, where $X>1$ such ...