29
votes
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 ...
15
votes
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 ...
11
votes
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 ...
10
votes
volume over a hypercube, over simplex: twist by Euler numbers
This is only a partial answer. The Beukers-Kolk-Calabi change of variables
$$x_1=\frac{\sin{u_1}}{\cos{u_2}},\;\;x_2=\frac{\sin{u_2}}{\cos{u_3}},\ldots,
\;x_{n-1}=\frac{\sin{u_{n-1}}}{\cos{u_n}},\;\;...
8
votes
Accepted
How to integrate an exponential function of an exponential function?
You can expand the integrand in powers of z. The coefficient of $z^k$ is
$$e^{-ky^2}\sum_{j=1}^{k-1} {1\over j!(k-j)!}=e^{-ky^2}(2^k-2)/k!.$$
This yields the following series expression for the ...
8
votes
A system of non-linear equations that is decomposable as a product -- uniqueness of solution?
The linear-in-$g$ system is uniquely solvable whenever $f_0\ne0$. therefore the solution set is parametrized by $f$ in the form
$$g=f_0^{-10}P(f;a),$$
where $P$ is linear in $a$, polynomial in $f$ ...
7
votes
Accepted
Physical interpretation of the mellin transform variable?
Physical interpretation: To develop a physical intuition, this article might be informative:
The power spectrum of the Mellin transformation with applications to scaling of physical quantities
The ...
7
votes
Accepted
Asymptotic behaviour of an integral
With the aid of Mathematica I found that
$$g(\theta)=\textrm{EllipticK}(a^2)-\textrm{EllipticF}(t,a^2).$$
I get the first terms of the asymptotic expansion
$$\frac{\sqrt{\pi}}{2\sqrt{k}}-\frac{\sqrt{\...
7
votes
Do distance functionals separate probability measures?
On the positive side, the answer is affirmative if $\Omega$ is the unit interval $[0,1]$ with its standard distance. In this case $\phi_\mu$ is a convex $1$-Lipschitz function (in fact, it is also ...
6
votes
Uncertainty principle for Mellin transform
There is something the above posts missed (perhaps because the wording didn't make it explicit): the condition on the function $f(x)$ is one sided (i.e., it assumes something on the decay as $x\to \...
6
votes
When is an integral transform trace class?
A remark on (Q3):
There is this famous example of T.Carleman (1916 Acta Math link) where he constructs a (normal ) operator with a continuous kernel such that it belongs to all Schatten p-classes if ...
6
votes
Asymptotics of Fresnel integrals
This is a classical, and very rich subject. A few years ago I advised a senior thesis on this subject. It came out nicely. I think that this thesis is a good place to start. It also has ...
6
votes
Accepted
Geodesics in finite groups
First, my apologies for this late answer, I only found the question today.
Below, I probably recall too many things, but I felt it could put some context around the short answer to question 1 saying: ...
6
votes
Radon transform between affine grassmannians
Yes, it is known.
See Boris Rubin, Radon transforms on affine Grassmannians, 2004, in particular theorem 4.2.
A free full text pdf is available at the linked page.
The reconstruction formula is not ...
6
votes
Accepted
Asymptotic behavior of integral with gamma functions
If I just insert the large-$z$ asymptotics of $\Gamma(z)\rightarrow \sqrt{2 \pi } e^{-z} z^{z-\frac{1}{2}}$, and take $z>1/2$ real for simplicity, I find
$$5^z\,F(z)\rightarrow \int_0^\infty \left(...
6
votes
Why Mellin transform is omitting infinite terms?
This question has not been well received, but I am intrigued by the delta function Mellin transform and would like to respond. I found this 2004 paper by Norbert Südland and Gerd Baumann instructive. ...
6
votes
Accepted
Injectivity of a class of integral operators
If the support of $\mu$ is contained in $[0, b]$ for some $b \in (0,1)$, then $T_\mu f = 0$ for any $f$ that is $0$ on $[0,b]$, so it is not injective.
EDIT: Another example, where the support of $\...
6
votes
Integral with 4 Bessel functions and an exponential
Let's consider the second integral, which can be written in the following form:
$$
I(p, q, i, j, k, l; a, b, c, d)
:=
\int_0^\infty
dt\,
\exp(-p t^2)
t^q
j_i(a t) j_j(b t) j_k(c t) j_l(d t)
$$
where ...
5
votes
Accepted
Variations on the Mellin and Dirichlet transforms
I know that the discrete Mellin transform was defined by V.S.Ryko:
http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=ivm&paperid=5138&option_lang=rus
English reference: Soviet ...
5
votes
Accepted
Infinite dimensional version of a simple fact on certain singular matrices
For the first question, the answer is not necessarily.
Very rough idea: The rank-nullity theorem doesn't always hold on infinite dimensional spaces.
Rough idea: Let the operator $A$ be defined on $...
5
votes
Accepted
An integral identity relate to the Gamma function or the Beta function
To evaluate $\int_{-\infty}^\infty {dx\over (1+ix)^a\,(1-ix)^b}$, view this as $\int_{-\infty}^\infty \widehat{f_a}(x)\,\overline{\widehat{f_{\overline{b}}}(x)}\,dx$ where $f_c$ are functions whose ...
5
votes
Relationship between the Radon transform and Twistor spaces
The correspondence between the Radon transform (from a space of real-valued functions on $\mathbb{R}^2$ to the space of functions on the manifold of straight lines in $\mathbb{R}^2$) and the Penrose ...
5
votes
2D Fourier transform of log function
If $v(r)=\log r$, then $\Delta v=\delta$ and $\hat {(\Delta v)}=1$, that is $-(\xi^2+\eta^2)\hat {v}(\xi, \eta)=1$. However this yields $-\hat{v}=1/(\xi^2+\eta^2)$ which has no sense since this ...
5
votes
Accepted
Fast computation of convolution integral of a gaussian function
Convolution with a Gaussian kernel of an $n$-point function has $n^2$ complexity, while Fourier transformation (FFT), multiplication, and inverse Fourier transformation is only of complexity $n\log n$....
5
votes
Solution to $\int_{0}^{y} x^{-a} \exp \left[- \frac{(b - cx^{-d})^2}{2} \right] dx$
Maple does not know a symbolic answer for this. The very special case $a=2,b=0,c=1,d=2$ is evaluated by Maple in terms of the Whittaker M function or a $\;{}_1F_1$ hypergeometric function:
$$
\int_{0}...
5
votes
Accepted
Is there a solution to $\int_{\theta-1}^{x} \left(\frac{y+1-\theta}{\theta} \right)^{-\frac{1}{\epsilon+3}} y^{-\frac{\epsilon+4}{\epsilon+3}}dy$?
With some effort (the lower integration limit requires care) I found this answer for the definite integral:
$$I(x)=\int\limits_{\delta}^{x} \left(y-\delta\right)^{-\frac{1}{ \epsilon+3}} y^{-\frac{\...
5
votes
How to solve the following $0= \int_{-\infty}^\infty e^{-\frac{(bt+\omega)^2}{2}} f(t+\omega) \frac{1}{i t} dt, \forall \omega \in \mathbb{R}$
It may be helpful to rewrite this in a way that avoids the principal value:
$$0=\int_{-\infty}^\infty e^{-(bt+\omega)^2/2} f(t+\omega) \frac{2}{i t} dt=\int_{-\infty}^\infty dt \int_{-\infty}^{\infty}...
4
votes
Accepted
Interpretation of the integral "with respect to a plane wave" in terms of Radon transform
Sure you can't - but somehow you can. Obviously, $x\mapsto h(\theta\cdot x)$ is not an integrable function (if not $\equiv 0$) since it is constant along lines perpendicular to $\theta$. However, if $...
4
votes
Accepted
Is it possible to get an equation with two exponentials and a bessel function in closed form?
Maybe you can accept the following result as a closed form. Transforming the variable of integration in your integral
$$
I_{n}(a,b):= \frac{1}{2 a}\ e^{-a/2}\int_{0}^{\infty} dx\ x^{(b-2)/4} e^{-x/2}(...
4
votes
Techniques to solve equations involving a definite integral
The problem space of symbolic computation on definite integrals is currently fairly open. The Risch algorithm answers the question if there is a closed form solution to a indefinite integral (assuming ...
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