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7

I changed my evaluation slightly, and I was able to get the result in a very simple form. First notice that $$ \int_{-\infty}^{\infty} \text{Ei}^{2}(-|x|) e^{ikx} \, dx = 2 \int_{0}^{\infty} \text{Ei}^{2}(-x) \cos(kx) \, dx. $$ Then integrating by parts, and assuming for now that $k >0$, $$ \begin{align}\int_{-\infty}^{\infty} \text{Ei}^{2}(-|x|) ...


3

Let the Fourier transform be defined as \begin{align} f(\omega) = \frac{1}{2\pi} \, \int_{-\infty}^{\infty} f(x) \, e^{-i \omega x} \, dx \end{align} Now, for $f(x) = e^{- x^{2} + 2x}$ the following is developed. \begin{align} 2 \pi \, f(\omega) &= \int_{-\infty}^{\infty} e^{- (x^2 - (2 - i\omega) x)} \, dx \\ &= e^{-\left(1 - \frac{i ...


2

I propose another method that does not use integration. Problem is in the change of variable when completing the square : because of the imaginary unit $i$, the new variable is complex and thus the integral must be done carefully in $\mathbb{C}$. As $f$ is smooth (i.e. $f\in\mathcal{C}^\infty\left(\mathbb{R};\mathbb{R}\right)$), $\hat{f}$ is also smooth. ...


2

Well, derivation under the integral sign gives you directly $$\delta(y)=\int e^{-ixy}dx$$ $$\delta'(y)=-i\int xe^{-ixy}dx$$ $$\delta''(y)=-\int x^2 e^{-ixy}dx$$ $$\delta'''(y)=i\int x^3 e^{-ixy}dx$$ Then, by integration by parts, you can verify what it does as a distribution: $$\int \delta(y)f(y)dy=f(0)$$ $$\int ...


2

You have $g=F(f)$. The Fourier transform is a continuous linear map in many good spaces (Schwartz space, its dual, or $L^2$). The derivative of a linear map is the map itself, and so a continuous linear map is continuously differentiable (and the second derivative vanishes). That is, $\partial g/\partial f=F$. (One usually denotes this by $DF(f)=F$ or ...


2

To elaborate on Daniel Fischer's comment: We have Let $$\tag1 f\colon x\mapsto \sum_{n=0}^\infty a_nx^n$$ be a power series and $R=1/\limsup\sqrt[n]{|a_n|}>0$ its radius of convergence. Then the series converges absolutely for all $x$ with $|x|<R$ and we can take the derivative termwise, i.e., $f'(x)=\sum_{n=0}^\infty (n+1)a_{n+1}x^n$ where ...


2

Consider $f(x) = e^{-\frac{1}{(1 - x^2)}}$ Then note that if $f^{n}(x) = \frac{P(x)}{Q(x)}f(x)$ then $f^{n+1}(x) = \frac{P'(x) Q(x) - P(x)Q'(x)}{Q(x)^2} + (-\frac{1}{1 - x^2})'f(x) = \frac{\hat{P}(x)}{\hat{Q}(x)}f(x)$. Where $f^n$ is the n th derivative of $f$ and $P,Q, \hat{P}, \hat{Q}$ are polinomials. Note that $Q(x) = \bigg(\frac{1}{1-x^2}\bigg)^k$ for ...


1

Try in the opposite direction: \begin{align} \mathcal{F}(e^{-ax}H(x)) &=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-ax}H(x)e^{-isx}dx \\ & = \frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-ax}e^{-isx}dx \\ & = \frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-x(a+is)}dx \\ & = ...


1

This answer is intended to complement David's very nice solution by showing that the partial sums of the Fourier integrals are at most be pointwise unbounded almost everywhere (a.e.), when $f\in L^{1}(\mathbb{R})$; there exists a subsequence which converges to $t$ pointwise a.e. Combining this with David's solution, we see that the partial sums of the ...


1

Ok. Note that $2\pi=1$ here... Say $f$ is Kolomogorov's example. Regard $f$ as a periodic function defined on the entire line. Let $c_n$ be the $n$-th Fourier coefficient of $f$ (we want to reserve \hat for the Fourier transform). Say $\psi\ne 0$ is a smooth function with support in $(-1/2,1/2)$. Say $\psi$ is even just so we don't have to worry about the ...


1

$g(x)=\sum\limits_{n=1}^\infty \frac{\alpha^n}{sin(x)}(sin(n-2)x cos2x + cos(n-2)x sin2x)$ $ = \sum\limits_{n=1}^\infty \alpha^n\frac{sin(n-2)x}{sin(x)} + 2\sum\limits_{n=1}^\infty \alpha^n cos(n-1)x$ $= -\alpha + \alpha^2\sum\limits_{n=1}^\infty\alpha^n \frac{sin(nx)}{sinx} + 2\sum\limits_{n=1}^\infty\alpha^n cos(n-1)x$ where we can find $g(x)$ to be: ...


1

I don't think so - it looks like you are missing a factor of $n$. I think it's a little simpler than this if you stick with exponentials. Write $$f(x) = \sum_{n=-\infty}^{\infty} 3^{-n^2} e^{i n x} $$ This converges, as you said, absolutely and uniformly. We may write $$f'(x) = i \sum_{n=-\infty}^{\infty} n \, 3^{-n^2} e^{i n x} $$ $$g(x) = \pi ...


1

An idea: At $\;\xi=0\;$ : $$\lim_{\xi\to0}f(\xi)\stackrel{l'Hospital}=\lim_{\xi\to0}\frac{i\lambda e^{i\lambda\xi}}i=\lambda$$ so that $\;\xi=0\;$ is a removable singularity. At any other point $\;f(\xi)\;$ is the quotient of two analytic functions and is thus analytic.


1

You forget to mention one vital thing: $f$ is periodic with periodicity $2\pi$. The fourier coefficient is as you mention: $$f_n=\frac{1}{2\pi}\int_0^{2\pi}f(x)e^{-inx}dx$$ An interesting property of this integral is that one may move the interval of integration: as $f(x+2\pi)=f(x)$ and also $e^{-in(x+2\pi)}=e^{-2n\pi i} \cdot e^{-inx}=e^{-inx}$, we may ...


1

Your solution is correct, apart from the fact that you forgot the scale factors for the Dirac terms. It should be $$\hat{T}(k)=\frac{1}{\sqrt{2\pi}}\left( \frac{2i\beta -\lambda k-\mu ik^2}{k^3}\right)+\sqrt{\frac{\pi}{2}}\left( \mu\delta(k)+i\lambda\delta'(k)-\beta\delta''(k)\right)$$ It looks like Mathematica can't be trusted here. First of all, it gets ...


1

Define $$f(t)=\sum_{n=-\infty}^\infty \hat\mu(n)e^{int}.$$The series converges uniformly so $f$ is continuous. The uniform convergence also shows that $$\hat f(n)=\hat\mu(n).$$So uniqueness (for complex measures) shows that $\mu=f$, or more carefully $d\mu=f\,dt$. If you don't buy the uniqueness for complex measures bit: Lemma: If $\nu$ is a complex ...



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