All Questions
9 questions
0
votes
1
answer
115
views
Fourier transform of exponential over torus
I found the following formula for the Fourier transform on a flat 2-torus, but I don't quite know how to derive it. We have a variable $q=(q_x,q_y) \in [0,2\pi)^2$ and by considering it in polar ...
- 128k
9
votes
1
answer
429
views
A curious norm related to the L¹ norm
If $f \in C^0([0,1])$, one can define
$$\Vert f \Vert_? = \sup_{J \subset [0,1]} \left\lvert \int_J f \right\rvert,$$
where $J$ runs among all subintervals of $[0,1]$.
This is a norm on $C^0([0,1])$ (...
CommunityBot
- 1
5
votes
0
answers
243
views
Is there a way to solve this integral on the sphere explicitly?
Let $k_{j}\in {\mathbb{Z}}^{+}$ and $\,a_{j}\in \,]0,1[$, be such that
$k_{j}\,a_{j}<1$, $j=1,\cdots,n$. Let $f:\mathbb{R}^{n}\rightarrow [0,\infty[$ be defined by the integral
$$f(y):=\int_{\...
- 852
0
votes
0
answers
88
views
Closed formula for iterated Fourier series
I'm trying to obtain a closed formula for the following integral.
\begin{align}
I_n = {} & \int_0^h \Bigr[\sum_{r_1=1}^\infty a_{1,r} \cos\left(\frac{2\pi}{h} r_1t_1\right) \\[6pt]
& {}+ b_{1,...
- 293
18
votes
3
answers
3k
views
A curious sin-integral
While contending with a certain Fourier series, I stumbled on an incredibly simple evaluation (numerically) of a slightly complicated-looking sin-integral.
So, I wish ask:
Question. Is this really ...
- 43.2k
4
votes
1
answer
191
views
Scaling of double convolution
I am interested in the scaling of
$$F(x_1,x_4)=\int_{\mathbb R^2} e^{-\vert x_1 -x_2 \vert -\varepsilon \vert x_2 -x_3 \vert- \vert x_3 -x_4 \vert } \ dx_2 dx_3 $$
In particular, I suspect that
$$F(...
- 128k
2
votes
1
answer
295
views
Non-zero, bounded, continuous, differentiable at the origin, compactly supported functions with everywhere non-negative Fourier transforms
Do there exist functions $F(x) \! : \, \mathbb R \to \mathbb R$ which are non-zero and bounded:
$$ \mathrm {Range} (F) = [l, u] \, , \quad \mathrm {where} \quad l, u \in \mathbb R \land u > l \, ; \...
- 179
10
votes
1
answer
833
views
This is not a dyadic cosine-product
The double-angle formula, $\sin2x=2\sin x\cos x$, turns the scary-looking integral
$$\int_0^{\infty}dz\prod_{k=1}^{\infty}\cos\frac{z}{2^k}$$
into fun once you realize $\prod_k\cos\frac{z}{2^k}=\frac{\...
- 105k
6
votes
1
answer
741
views
Is the following integral nonzero?
Recently I met an integral as follow:
$$\int_0^{2\pi}\cdots\int_0^{2\pi}\left(\prod\limits_{1\leq i<j\leq9}\sin\frac{\theta_i-\theta_j}{2}\right)\left(\prod\limits_{i=1}^9(1+\cos(\theta_i-\theta_{i+...
CommunityBot
- 1