16

In fact, precisely your integral has been computed in closed form, in: Annie Gervois and Henri Navelet, Some integrals involving three Bessel functions when their arguments satisfy the triangle inequalities, J. Math. Phys. 25 (1984), no. 11, 3350–3356. Their result is
$$
\int_0^\infty J_m(ar)J_n(br)J_{m+n}(cr)r\,dr =
\begin{cases}
0&\text{if }c^2 < (a-...

answered Jan 15 '14 at 22:43

Francois Ziegler

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12

Combining Steve's idea with the Beurling-Hormander uncertainty principle yields the kind of result you're looking for.
The BHUP is the following:
Beurling-Hormander Uncertainty Principle: If $f\in L^1(\mathbb{R})$ and $$ \int\int_{\mathbb{R}^2}|f(x)\hat f(y)|e^{|xy|}dxdy<\infty, $$ then $f=0$.
As a corollary, we have a Hardy uncertainty principle ...

11

There is a simple proof along the following lines: Because the first deRham cohomology group of $\mathbb{RP}^2$ is trivial, a $1$-form on $\mathbb{RP}^2$ is exact if and only if it is closed. Let $\alpha$ be a $1$-form on $\mathbb{RP}^2$ whose integral over every line vanishes, and let $\beta = \pi^*\alpha$ where $\pi:S^2\to \mathbb{RP}^2$ is the standard ...

answered Jan 2 '13 at 20:26

Robert Bryant

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10

The growth rate is exponential. Specifically, I will prove the bound $J_n \leq \frac{(n+1)^{n} 2^n}{(n+1)!}$, which grows like $(2e)^n$. I can also show that $J_n \geq 2^n$. I would guess that neither of these are the true rate of growth.
I'll also give an explicit recursion for the $J_n$, which makes it clear that they are all in $\mathbb{Q}[\pi]$ and ...

answered May 23 '13 at 22:31

David E Speyer

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10

This is only a partial answer. The Beukers-Kolk-Calabi change of variables
$$x_1=\frac{\sin{u_1}}{\cos{u_2}},\;\;x_2=\frac{\sin{u_2}}{\cos{u_3}},\ldots,
\;x_{n-1}=\frac{\sin{u_{n-1}}}{\cos{u_n}},\;\;x_n=\frac{\sin{u_n}}{\cos{u_1}}$$
has the Jacobian
$$\frac{\partial(x_1,\ldots,x_n)}{\partial(u_1,\ldots,u_n)}=
1-(-1)^n\,x^2_1x^2_2\cdots x^2_n.$$
Therefore ...

9

Here is one solution to compute $\int_0^1 \frac{1}{\sqrt{1-x^4}}dx$, with a big accuracy, without using arithmetic-geometric means.
The variable change x = sin u gives
$\int_0^1\frac{1}{\sqrt{1-x^4}}dx = \int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{1+sin^2 u}}du$
The second integral can be computed quickly using Simpson's rule, 8 intervals are enough for eleven ...

9

This triple-Bessel integral was studied in much detail by S.K.H. Auluck, On the Integral of the Product of Three Bessel Functions over an Infinite Domain. Closed form expressions are given if two the three orders coincide. For the more general case, a reliable way to evaluate this integral using Mathematica is described and tested. The numerical difficulties ...

answered Jan 13 '14 at 5:37

Carlo Beenakker

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8

Two canonical online references are:
The Digital Library of Mathematical Functions (http://dlmf.nist.gov/)
This is the official successor of the venerable Handbook of Mathematical Functions by Abramowitz and Stegun.
The Wolfram Functions Site (http://functions.wolfram.com/)
An expansive collection of identities and properties of special functions amassed ...

8

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 integral:
$$\sum_{k=1}^\infty \sqrt{\pi\over k}(-2+2^k)z^k/k!.$$

ca.classical-analysis-and-odes sequences-and-series integration stochastic-calculus integral-transforms

8

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 $f$ and homogenenous of degree $9$. Not only it is a continuum, but it has dimension $10$.
When $f_0=0$, there is no solution in general, say for $a_1\ne0$.

7

A good place to look is pages 405 and 413 of the Nachlass section of Gauss's Werke III, which can be found online through Google Books. On page 405, he gives the following formula for "$\text{arc sin lemn }x$":
$$\text{arc sin lemn }x= x+{1\over2}\cdot{1\over5}x^5 + {1\cdot3\over2\cdot4}{1\over9}x^9+{1\cdot3\cdot5\over2\cdot4\cdot6}{1\over13}x^{13}+{1\...

7

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 Mellin transform is used to diagonalize the dilation operator in a
manner analogous to the use of the Fourier transform to diagonalize
the translation ...

answered Sep 4 '15 at 13:44

Carlo Beenakker

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6

David's approach is, indeed, superb. Here is an alternative proof with more analytic flavor. It gives a worse constant as presented here (I made next to no attempt to optimize it), but it shows a bit more and can also teach you a couple of useful techniques.
First, let's clear the ground a bit. The problem is equivalent to estimating the integral
$$
\...

6

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 and only if $p\geq 2.$
More precisely it's possible to construct $k(x)=\sum_n c_ne^{2\pi i n x}$ continuous and periodic with $\sum_n|c_n|^p=\infty$ for $p<...

6

If my chalkboard scribblings are correct, if $f(x) = \cos(\alpha x)$ and $g(x) = \sin(\alpha x)$ for $\alpha \neq 0$, then
$$ \mathscr{T}_k f(x) = \alpha^{-2}(f(x) + \alpha x\sin(\alpha T) - 1), \quad \mathscr{T}_k g(x) = \alpha^{-2}(g(x) - \alpha x \cos(\alpha T))$$
(up to some inconsequential signs), so that your orthonormal basis of eigenfunctions is
$$ \...

6

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 $x\to \infty$, not as $x\to -\infty$), and thus, once we take logarithms to make the Mellin transform into a Fourier transform, we end up with a one-sided condition as ...

6

define $G(y)=y^{-1}\exp(y^2/2)(Tf)(y)$ and $g(x)=x^{-1}\exp(-x^2/2)f(x)$, then you seek the solution to the integral equation
$$G(iq)=\int_{0}^{\infty}g(x)J_{0}(xq)xdx$$
This is a Hankel transform. The inverse is
$$g(x)=\int_{0}^{\infty}G(iq)J_{0}(xq)qdq$$

answered Oct 10 '13 at 22:09

Carlo Beenakker

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6

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: yes, under the names "geodesic in an involutory quandle", or "cycle in a symmetric set".
1) Recall that a homogeneous symmetric space (I borrow the terminology ...

6

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 many useful references. Most of what you need is in Section 2.

answered Apr 8 '15 at 15:16

Liviu Nicolaescu

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6

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
$$\frac{\sqrt{\pi}}{2\sqrt{k}}-\frac{\sqrt{\pi} x^2}{8(1-a^2)k^{3/2}}+
\Bigl(-\frac{k}{6}+\frac{a^2 x^2}{6(1-a^2)^2}+\frac{x^4}{24(1-a^2)^4}\Bigr)
\frac{3\sqrt{\pi}}{8k^{5/2}}$$
For example with $a=2/3$,...

6

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
$$5^z\,F(z)\rightarrow \int_0^\infty \left(1+x^2/z^2\right)^{z-\frac{1}{2}} e^{\pi x/2-2 x \arctan \left(x/z\right)}\,dx$$
$$\qquad = z\int_0^\infty(1+x^2)^{-1/2}\exp[z u(x)]\,dx$$
$$\text{with}\;\;u(x)=...

answered Apr 15 '17 at 17:38

Carlo Beenakker

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6

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. The Mellin transform pair is
$${\cal M}[f(x),s]=\int_0^\infty f(x)x^{s-1}\,dx$$
$${\cal M}^{-1}[g(s),x]=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}g(s)x^{-s}\,...

answered Jul 9 '17 at 18:53

Carlo Beenakker

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6

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 $\mu$ is all of $[0,1]$: $d\mu(x) = g(x)\; dx$ where
$g(x) = 5/3$ for $0 \le x < 1/2$, $1/3$ for $1/2 \le x \le 1$. Let $f(x) = x \sin(\pi \log_2(x))$, and ...

5

Gradshteyn and Ryzhik
http://www.amazon.com/Table-Integrals-Series-Products-Edition/dp/0122947576/ref=sr_1_1?ie=UTF8&qid=1339111210&sr=8-1
is "the bible" for many people. At one point I owned two copies. It summarizes results from the Bateman project, with very precise references. It also has culled identities from many other sources. In a fairly ...

5

you're missing a factor $N$ because $\rho(\lambda_1,...,\lambda_j,...,\lambda_N)$ is normalized to unity, while $\rho_{(1)}(\lambda)$ is normalized to $N$:
\begin{align}
\int_{-\infty}^\infty d\lambda_1... \int_{-\infty}^\infty d\lambda_j\ ...\int_{-\infty}^\infty d\lambda_N\ \frac{\rho(\lambda_1,...,\lambda_j,...,\lambda_N)}{\bigl(\lambda_j-(\lambda+i\...

answered Dec 4 '13 at 15:09

Carlo Beenakker

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5

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 Mathematics (Izvestiya VUZ. Matematika), 1991, 35:8, 63–66
He also developed a very strong method with many page tables to sum series based on it (reference [4] in ...

5

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 very simple, so I will not reproduce it here.
Radon transforms on Grassmannians are also discussed in Helgason's book Radon Transform, section II.4.F.

answered Aug 15 '15 at 21:04

Joonas Ilmavirta

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5

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 $L^2(M)$ be a injective mapping such that its range does not include the constant function. More precisely, since $M$ is compact we can enumerate its ...

5

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 Fourier transforms are $(1+ix)^{-c}$. Use the Gamma-function identity
$$
\int_0^\infty e^{-ty}\,t^c\;{dt\over t}\;=\; y^{-c}\cdot \Gamma(c)
$$
at first for real $...

5

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 transform (from $\mathbb{CP}_2$ to the dual projective space $\mathbb{CP}^\ast_2$) is easiest to work out when lines in the Radon transform are replaced by great ...

answered May 22 '17 at 11:57

Carlo Beenakker

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