# Tag Info

We wish to evaluate $$\int_{-\infty}^{\infty} e^{-|x|^{\alpha}}e^{-2\pi ixy}\,dx.$$ Making use of evenness, we get $$2\int_0^{\infty} e^{-x^{\alpha}} \cos(\pi xy)\,dx.$$ Expanding $\cos$ in a power series, we have $$2\sum_{n=0}^{\infty}\frac{(-1)^n\pi^{2n} y^{2n}}{(2n)!}\int_0^{\infty} e^{-x^{\alpha}} x^{2n}\,dx.$$ So we need only to evaluate the ...
This equation $$f'(x)=2f(2x+1)-2f(2x-1), \quad f(0) = 1, \tag{*}$$ has a finite solution which is also known as the $\mathrm{up}(x)$ or $\mathrm{hut}(x)$ function. It has compact support $\mathrm{supp}\,\mathrm{{up}}(x)=[-1,1]$ and its Fourier transform is $\hat{f}(t)=\prod\limits_{k=1}^{\infty}\mathrm{sinc}{(t\cdot ... 2 There is no Schwartz space, just$C^\infty(\mathbb T^n)$. Indeed,$\mathcal S(\mathbb R^n)$is an intermediate class between all smooth functions and compactly supported smooth functions. But on$\mathbb T^n$, everything is compactly supported. Since the Fourier transform of a function on$\mathbb T^n$is a function on$\mathbb Z^n$, it does not make ... 2 For the$s=0$case, let$s_{n}=\sum_{k=1}^{n}c_{k}$for$n \ge 1$and let$s_{0}=0$. Then$s_{n}-s_{n-1}=c_{n}$for all$n \ge 1$and$\lim_{n} s_{n}=0$by assumption. So the following are absolutely convergent for$0 \le r < 1: \begin{align} \sum_{n=1}^{\infty}r^{n}c_{n} & = \sum_{n=1}^{\infty}r^{n}(s_{n}-s_{n-1}) \\ & = ... 2 Do not use the triangle inequality to bring the absolute value inside the integral — it will not help, as \lvert e^{-inx}\rvert =1. Instead, use integration by parts. \int_{ - \pi }^\pi nf (x)e^{- inx} dx = \left[i f(x)e^{- inx}\right]^\pi_{-\pi} -i \int_{[-\pi,\pi]} f^\prime(x) e^{- inx}dx $$For the first term, leverage the periodicity of f. For ... 2 Edit: This is the answer for the first version of the question. If period means the smallest period the answer is: no. Constant functions are counterexamples. 2 If you have a linear combination of functions, the resulting Fourier series is the corresponding linear combination of the Fourier series of the functions. So yes, it is linear. The key is that it is linear in the coefficients, even though the series is not linear in x. 2 If we take out the 2 \pi for simplicity's sake and rescale time such that f_m/f_c=r and f_c=1, then we are basically considering$$s(t)=\sin((1+\sin(rt))t) = \sin(t+t\sin(rt))$$Now consider r=1 and going forward in time by 2 \pi:$$s(t+2 \pi)=\sin(t+2 \pi + (t+2\pi) \sin(t+2\pi)) \\ = \sin(t + (t+2 \pi) \sin(t+2 \pi)) \\ = \sin(t + (t + 2 \pi) ... 1 For a functionf(x)=\frac{1}{\pi}\frac{a}{a^2+x^2}$(also kwnown as Lorentzian function) the Fourier transform is: $$\int_{-\infty}^{\infty}\frac{1}{\pi}\frac{a}{a^2+x^2}e^{-itx}dx=\frac{a}{\pi}\int_{-\infty}^{\infty}\frac{1}{(a+ix)(a-ix)}e^{-itx}dx$$ Now we will use Cauchy's integral formula. Although we only need to solve our integral for the real axis, ... 1 In your Calculation, x shouldn't be the lower border in the integral. It must be H(x)*x. Because, if x < 0, the integral is cut, then the lower border is 0. And in the end it is (x-2H(x)*x) = -|x| ... :) 1 Yes, the Fourier Series is linear. The coefficients$a_k$and$b_k$are defined in terms of integrals and integrals are linear. For example, for constants$\lambda$and$\mu$and function$\mathrm{f}$and$\mathrm{g}$we have $$\frac{1}{L}\int_{-L}^L \left(\lambda\mathrm{f}(x)+\mu\mathrm{g}(x)\right)\cos\left(\frac{\pi n x}{L}\right) \mathrm{d}x=$$ ... 1 Suppose the Fourier series of a periodic function$f$with period$2\pi$is $$a_0+\sum_{k=1}^\infty \left(a_k \cos kx+b_k \sin kx\right)$$ and the Fourier series of$g$is $$q_0+\sum_{k=1}^\infty \left(q_k \cos kx+r_k \sin kx\right).$$ Then the Fourier series of$mf+ng$, where$m$and$n$are constants, is $$(ma_0+nq_0)+\sum_{k=1}^\infty \Big((ma_k + ... 1 We have:$$ \frac{\pi}{2}\,a_n=\int_{0}^{\pi}\sqrt{1-k^2\sin^2\theta}\cos(2n\theta)\,d\theta =\frac{k^2}{4n}\int_{0}^{\pi}\frac{\sin(2\theta)\sin(2n\theta)}{\sqrt{1-k^2\sin^2\theta}}\,d\theta.$$If we set:$$ b_m = \int_{0}^{\pi}\frac{\cos(2m\theta)}{\sqrt{1-k^2\sin^2\theta}}\,d\theta,\qquad c_m = \int_{0}^{\pi}\cos(2m\theta)\sqrt{1-k^2\sin^2\theta}\,d\theta ... 1 We denote by$\phi$the characteristic function of the random variable$X1_A/\mathbb{P}(A)$. Then we have $$\phi(t)=\mathbb E\left(\mathbb 1_A\exp\left(it\frac X{\mathbb{P}(A)} \right)\right)+1-\mathbb{P}(A).$$ I'm not sure that this can be simplified more or written as a function of the characteristic function of$X$and the indicator function of$A$. ... 1 Let us first establish the case$I=\left[-1,1\right]$. This yields$\lambda\left(I\right)=2$and$c\left(I\right)=0$. Now, we use a dyadic partitition of$\mathbb{R}\$, i.e. we write \begin{eqnarray*} \int\frac{\left|f\left(x\right)\right|}{\left(1+\frac{\left|x-c\left(I\right)\right|}{\lambda\left(I\right)}\right)^{N}}\,{\rm d}x & \lesssim & ...