Compute the Fourier transform of the following tempered distributions on $\mathbb R^n$, which, for $\phi \in \mathcal S(\mathbb R^n)$ are given by    $T_1(\phi) := \int_{\mathbb R^n} x^{\alpha} \phi(x) dx$, $\alpha \in \mathbb N_0^n$.     $T_2(\phi) := \int_{\mathbb R^n}\sin(\langle a, x \rangle_{\mathbb R^n}) \phi(x)dx$, $a \in     \mathbb R^n$.     $T_3(\phi) := \int_{\mathbb R} \mathbb 1_{[-1,1]}(x) \phi(x) dx$.   Here's what I know and what I've tried so far $\mathcal F$ denotes the Fourier transform $$ \mathcal{F} f (\xi) := \hat{f}(\xi) := (2 \pi)^{-n/2} \int_{\mathbb R^n} f(x) e^{-i \langle x, \xi \rangle} dx $$ and $\mathcal{S}$ the Schwartz space. We defined $(\mathcal{F} T)(\phi) = T(\mathcal{F}(\phi))$. I am also going to use the following rules for the Fourier transform: $D^{\alpha}(\mathcal{F} \phi)(\xi) = (-i)^{|\alpha|} \mathcal{F}(x^{\alpha} \phi)(\xi)$ and $\xi^{\alpha} (\mathcal{F}\phi)(\xi) = (-i)^{| \alpha |} \mathcal{F}(D^{\alpha} \phi)(\xi)$.  Is the following calculation correct? \begin{align*}             \mathcal{F}(T_1(\phi))             &amp; = T_1(\mathcal{F}(\phi))             = \int_{\mathbb{R}^n} x^{\alpha} \hat{\phi}(x) dx             = (-1)^{| \alpha |} \int_{\mathbb{R}^n} (\hat{\phi}(x))^{(\alpha)} dx \\             &amp; = (-1)^{|\alpha|} \int_{\mathbb{R}^n}  (-i)^{|\alpha|} \mathcal{F}(t^{\alpha} \phi)(x) dx             = i^{|\alpha|} \int_{\mathbb{R}^n} \mathcal{F}(t^{\alpha} \phi)(x) dx \\             &amp; = i^{|\alpha|} \mathcal{F}^{-1}(\mathcal{F}(t^{\alpha} \phi))(0)             = i^{| \alpha |} 0^{\alpha} \cdot \phi(0)             = \begin{cases} i^{| \alpha |} \phi(0), &amp; \alpha = 0, \\             0, &amp; \text{else.} \end{cases}         \end{align*}  I am pretty stuck with this one as I haven't found a way to further reduce this expression. Any hints? \begin{align*}         \mathcal{F}(T_2(\phi))         = T_2(\hat{\phi})         = \int_{\mathbb R^n} \sin(\langle a, x \rangle) \hat{\phi}(x) dx     \end{align*} The next step I will try is to shift the Fourier transform to the sin, as I have done in 3. But I am even struggling to evaluate the integral for $n = 1$, which is $$ \int_{\mathbb R} \sin(a x) e^{-i x \xi} dx = \frac{e^{-i \xi x}(a \cos(ax) + i \xi \sin(ax))}{\sqrt{2 \pi} (a - \xi)(a + \xi)}\bigg|_{x = - \infty}^{\infty} $$  Here I redristribute the Fourier transform inside the integral, which should be valid by Fubini. Is it valid in this case? \begin{align*}         \mathcal{F}(T_3(\phi))         &amp; = T_3(\hat{\phi})         = \int_{\mathbb{R}} \mathbb 1_{[-1,1]}(x) \hat{\phi}(x) dxf         = \int_{\mathbb{R}} \widehat{\mathbb 1_{[-1,1]}}(x) \phi(x) dx \\         &amp; = \int_{\mathbb{R}} \sqrt{\frac{2}{\pi}} \frac{\sin(x)}{x} \phi(x) dx     \end{align*}   Edit With @guy3141's hint I proceeded as follows: With $\Im(x) = \frac{x - \bar{x}}{2i}$ we have \begin{align} \int_{\mathbb R^n} \sin(\langle a, x \rangle) e^{-i \langle x, \xi \rangle} dx &amp; = \frac{1}{2i} \int_{\mathbb R^n} \left(e^{i \langle a, x \rangle} - e^{-i \langle a, x \rangle}\right)e^{-i \langle x, \xi \rangle} dx \\ &amp; = \frac{1}{2i} \int_{\mathbb R^n} e^{-i \langle \xi - a, x \rangle} - e^{-i \langle \xi + a, x \rangle} dx. \end{align} Now we can use Fubini's theorem (F) to evaluate both summands separately: \begin{align} \int_{\mathbb R^n} e^{- i \langle \xi \pm a, x \rangle} dx &amp; = \int_{\mathbb R^n} \prod_{k = 1}^{n} e^{-i x_k (\xi_k \pm a_k)} dx  \overset{\text{(F)}}{=} \prod_{k = 1}^{n} \int_{\mathbb R} e^{-i x_k (\xi_k \pm a_k)} dx_k, \end{align} but I fear that I can't use Fubini's theorem as the integral on the RHS doesn't converge.
123766
Let $T$ be a tempered distributions, and let $S(x) = e^{i\alpha x}T(x)$, where $\alpha \in \mathbb{R}$.    Find a formula that relates the Fourier transforms $\hat S$ and $\hat T$. Find the Fourier transform $\hat f_\alpha$ of $f_\alpha (x) = \sin \alpha x$. Find $\lim_{\alpha \to \infty} \alpha \hat f_\alpha$ in the sense of (tempered) distributions.
96707
Let $f_n = \mathbb 1_{[-n,n]}$ for all $n \in \mathbb{N}$   Compute explicitly $f_n \star f_1$ for all $n \in \mathbb{N}$.  Show that $f_n \star f_1$ is the Fourier transform of $g_n = \frac{  \sin{(2\pi x)} \sin{(2 \pi nx)}}{\pi ^2 x^2}$  Show that $\| g_n \|_1 \to \infty$  Deduce that the Fourier transform is not surjective from $\mathcal{L}^1(\mathbb{R})$ to $C_0( \mathbb{R})$    Here, $C_0(\mathbb{R})$ is the space of continuous functions $f : \mathbb{R} \to \mathbb{R}$ that tends to $0$ at $-\infty$ and $+\infty$. I managed to compute 1) and proved 2). But I don't know how to prove 3). Moreover, I don't see how can 1), 2) and 3) prove 4). Can anyone help?
84451
I'm having trouble figuring out the Fourier series of $|\cos(x)|$ from $-\pi$ to $\pi$.  I understand its an even function, so all the $b_n$s are $0$  $$a_0 = \frac 2 \pi \int_0^\pi |\cos(x)|\,dx = 0$$  $$a_n = \frac 2 \pi \int _0^\pi  |\cos(x)| \cos(nx) \, dx = \frac 2 \pi \int_0^\pi \cos^2(x)\,dx.$$  since for all $j,k$ not equal the integral is zero.  so only $a_1$ remains. is this correct?   How would I evaluate $\sum_{n=1}^\infty (-1)^{n-1} /(4n^2 - 1)\  {}$?
20679
We know that the integral $$ I = \int_{-\infty}^{\infty}  e^{-x^2}\ \mathrm{d} x $$ can be calculated by first squaring it and then treat it as a $2$-dimensional integral in the plane and integrate it in polar coordinates. Are there any other ways to calculate it?  I know that we may use the relation $$\Gamma(x)\, \Gamma(1-x) = \frac{\pi}{\sin{\pi x}}$$ but this, in effect, is still taking the square. Well, after I write down the above text, I figure that maybe there is no way to calculate it without squaring, since, after all, the result contains a square root, and it seems no elementary function can &quot;naturally&quot; produce a square root of $\pi$ starting from natural numbers (though I don't know how to describe this more concretely; you are also welcome to comment on this point).  Nevertheless I still post this question in case there are some other ideas. EDIT: the Fourier transformation method at  appears kind of cheat to me, since the very proof of the Fourier transformation formula actually makes use of the value of the Gauss integral (at least in this wiki page ). Thank you.
90617
I've been reading up a bit on control systems theory, and needed to brush up a bit on my Laplace transforms. I know how to transform and invert the transform for pretty much every reasonable function, I don't have any technical issue with that. My only problem is that some theoretical things are not entirely intuitive.  This is my current understanding of the Fourier transform: We have an function, member of an infinite dimensional space, and we want to decompose it in terms of basis functions $e^{j\omega t}$. To do so, we project (analogously to taking the dot product) our function onto the basis functions, using the following definition of an inner product for the function space: $\langle f, g\rangle = \int_{-\infty}^{+\infty}f(t)\cdot g^*(t)dt$, which gives rise to the Fourier transform: $F(j\omega) = \int_{-\infty}^{+\infty} f(t)\ e^{-j\omega t}\,dt$. Reconstructing the original function from the coefficients times the basis functions gives $f(t) = \frac{1}{2\pi}\int_{-\infty}^{+\infty} F(j\omega)\ e^{j\omega t}\,d\omega$  Now, looking at the definition of the Laplace transform, it looks like it's trying to do the same thing using underdamped/overdamped sinusoids of the form $e^{\sigma t}e^{j\omega t}$, instead of pure sinusoids like the FT. This enables representing functions which don't vanish at infinity, but rather can diverge exponentially. However, it doesn't quite fit in the framework I laid above. First of all, the inversion integral confuses me. $\mathcal{L}^{-1} \{F(s)\}(t) = f(t) = \frac{1}{2\pi j}\int_{\sigma-j\infty}^{\sigma+j\infty}F(s)e^{st}\,ds$. Why am I free to do the integration for any value of σ, as long as it's bigger than the real part of the rightmost pole? If I plug  in $s = \sigma + j \omega, ds = j d\omega$, I can pull $e^{\sigma t}$ out of the integral, to get $\mathcal{L}^{-1} \{F(s)\}(t) = f(t) = \frac{e^{\sigma t}}{2\pi}\int_{\infty}^{+\infty}F(\sigma+j\omega)e^{j\omega t}\,d\omega$. It appears strange to me that the $e^{\sigma t}$ term has to be exactly compensated by the fact that the transform is being evaluated more to the right or to the left. Also, when analyzing systems using the Laplace transform, the main thing we're interested in is the location of its zeros and poles. Intuitively, I would expect that, since the function "blows up" at the poles, there would be large contributions from those amplitudes in the inversion integral if it passed near the poles. However, if it can pass arbitrarily far away from the poles, it doesn't make as much sense...  I'm sorry if I'm not explaining myself as well as I wanted to, but this isn't really that easy to express...  Cheers
2342
I need help to calculate the  Fourier transform of this funcion  $$b(x) =\frac{1}{x^2 +a^2}\,,\qquad a &gt; 0$$ Thanks.
74985
For a compactly supported distribution $f$, I want to show the non-surjectivity of the map $H:f \mapsto f*f \in \mathcal{E}'$.     My attempt: I have chosen to show that this is not the case by showing that $\text{image } (H)$ does not contain $ \delta_0'$. For instance I restrict myself to $d=1$.   Suppose that $H$ is surjective. Applying the Fourier transform on the convolution, I get $(\hat{f}(z)) ^2 = i z$. The theorem of Paley-Wiener-Schwartz guarantees that $(\hat{f}(z)) $ should be an entire function. However, as it is well-known, no such entire function exists:   If this were true, then $e^{i \pi /4}\hat{f}(z)$ would be the square root function, which cannot be well defined on $\mathbb C$, and even less so entire.    Q1: Is there another way of achieving the result (i.e. with a choice significantly different than $\delta_0'$, and with an approach different than the one using the Fourier transform)?  Q2: Can I generalize this without much effort in the case of $d \ge 1$ (probably using the tensor product)?    Edit: The source is .
77491
I recently found out that the expectation of a random variable $X$ in a probability space $(\Omega, \mathcal F,  \mathbb P)$, $\mathbb E(X)$, is just the term used in probability theory for the measure of the function $X$; i.e.:  $$ \mathbb E(X)=\mathbb P(X) $$  or, for a real-valued random variable $X:\Omega\to\mathbb R$:  $$ \mathbb E(X)=\int_\Omega X(x)\mathbb P(dx) $$  The thing is, I am also familiar with the definition of the expectation of a random variable $X:\Omega\to\mathbb R$ with probability distribution function $f$ to be:  $$ \mathbb E (X)=\int_\mathbb R xf(x)dx $$  Here, the integral is with respect to the Lebesgue measure on $\mathbb R$.    We can define the function $f$ in the following way: we have $F(x)=\int_\mathbb R f(x)dx$, where $F(x)=\mathbb P(X^{-1}((-\infty,x]))$.    I can't see any way to show that these two definitions are equal to one another.  The only transformation between integrals with respect to different measures that I know is that $\mu(f^{-1}(g))=\mu(g\circ f)$, which gives that $\mathbb E(g)=\mu_X(g)$, where $\mu_X$ is the image measure $\mathbb P\circ X^{-1}$, but that doesn't seem to help.  Why can we write $\mathbb E(X)=\int_\mathbb R xf(x)dx$?  Why are these two definitions consistent?
23958
I would like to work out the Fourier transform of the Gaussian function   $$f(x) = \exp \left(-n^2(x-m)^2 \right)$$  It seems likely that I will need to use differentiation and the shift rule at some point, but I can't seem to get the calculation to work. Does anyone have any advice?  By the way, I am using   $$\mbox{FT}(f)(k)=\int_{-\infty}^\infty f(x)e^{-ikx}\,\mathrm d x$$  as my definition of a Fourier transform. Many thanks.
9190
I'm working on the Fourier transform, but I don't know how to evaluate the integral:  $$I = \int_{ - \infty }^{ + \infty } {\frac{{\cos (ax)}}{{{x^2} + 1}}{e^{ - ibx}}\,\mathrm dx} $$
33926
Let $H$ be the inner product space = $\{f: \mathbb{R} \to \mathbb{C} \mid f \text{ is continuous and has period }2 \pi\}$ where the inner product is: $$\langle f,g \rangle = \int_{0}^{2\pi}f(t)\overline{g(t)} dt $$  How do I prove that for $n \in \mathbb{Z}$ and $e_n(t)=\dfrac{1}{\sqrt{2\cdot\pi}} e^{i n  t}$, $(e_n)_{n \in N}$ is a basis of $H$ (that it is orthonormal is easy).
39984
$\mathrm{Re}(i) = 0$, but the fourier transform of $f(x) = e^{-ix^2}$ is $g(\alpha) = \sqrt{\pi\over i}\times e^{i\alpha^2 \over 4}$, is it not?   Is there an easy to show that it is so, knowing the fourier transform of $h(x) = e^{-x^2}$? (It is $k(\alpha) = \sqrt\pi\times e^{-\alpha^2 \over 4}$.)  The non-unitary, angular frequency Fourier transform is taken into consideration.
23086
Let $u$ be a distribution on $\mathbb{R}^n$ with support = $\left\{0\right\}$. Then there exists $N$ such that $u$ has order $N$. Let $\chi\in C_0^{\infty}(\mathbb{R}^n)$ a smooth function with  $\chi(x) = 1$ for $0\leq |x|\leq 1 $,   $\chi(x)\in [0,1]$ for $1\leq |x|\leq 2$,  $\chi(x)=0$ for $|x|\geq 2$  Denote $ \chi(x/r) = k_r(x)$ for $r\in (0,1]$. For the case $N=0$ I want to show that there exists $c_1$ such that $$|\left\langle u,\phi \right\rangle|\leq c_1 |\phi(0)|  $$ for all $\phi \in C_0^{\infty}(\mathbb{R}^n)$. The idea would be to apply $u$ on $\phi = k_r\phi+(1-k_r)\phi$ and letting $r\to 0$.   So then I guess we can write $$ |\left\langle u,\phi \right\rangle| \leq |\left\langle u,k_r\phi \right\rangle| + |\left\langle u,(1-k_r)\phi \right\rangle |$$ But not really sure how this follows...  Moreover how can i show that $\left\langle u,\phi \right\rangle = \left\langle u,\phi \chi\right\rangle = \left\langle u,\chi\right\rangle\phi(0) $?
98581
I am reading up on the fourier series, and I keep seeing it as being defined as:  $$ f(\theta)= \frac{1}{2}a_0 + \sum_{n=1}^{\infty}(a_n \cos(n\theta) + b_n \sin(n\theta)) $$  where  $$ a_n = \frac{1}{\pi}\int_{0}^{2\pi}\cos(n\theta)f(\theta)d\theta $$  and   $$ b_n = \frac{1}{\pi}\int_{0}^{2\pi}\sin(n\theta)f(\theta)d\theta $$  I understand the derivation of the coefficients using trig integral identities, but I can't find a clear explanation of why $\frac{1}{2}$ is in front of $a_0$. Can anyone help show my why this is the case? Why can't we just have $a_0$ with no number in front of it. Thanks!  edit: corrected summation term
37770
Consider a function $f \in L^2(\mathbb{R}^n)$ such that $f$ is radial. My question is, is the Fourier transform $\hat{f}(\xi)$ automatically radial (I can see it is even in each variable $x_i$), or we need some conditions on $f$? Thanks for your help.
119625
I am trying to calculate  $$Z = \int\limits_{\phi(\beta) = \phi(0) =0} D \phi\ e^{-\frac{1}{2} \int_0^{\beta} d\tau \dot{\phi}^2}$$  without transforming it to the , I can show that $Z = \sqrt{\frac{1}{2\pi \beta}}$. However, I have a problem in obtaining the same result in the Matsubara frequency space: \begin{equation} \phi (\tau) =  \frac{1}{\sqrt{\beta}} \left( \sum_{n} \phi_n \ e^{i\omega_n\tau} \right), \end{equation} with $\sum_n \phi_n =0, \omega_n = \frac{2\pi n}{\beta}$. And \begin{equation} Z = \int \prod_n D\phi_n\ \delta\left(\sum_n \phi_n\right)\ e^{-\frac{1}{2} \sum_n \phi_n \phi_{-n}  \omega_n^2  }  \end{equation} which, I think, vanishes.   I guess the problem lies in the measure. Any comments?  Info: I write the Schulman's derivation in imaginary time here. \begin{eqnarray} Z &amp;=&amp; \int\limits_{\phi(0) =\phi(\beta) = 0} D\phi(\tau) e^{-\frac{1}{2}\int_0^{\beta}d\tau\dot{\phi}^2}\\ &amp;=&amp; \text{lim}_{N \rightarrow \infty} (\frac{1}{2\pi \epsilon})^{(N+1)/2} \int d\phi_1 \dots d\phi_N e^{-\frac{1}{2\epsilon} \sum_{i =0}^N (\phi_{i+1} -\phi_i)^2} \end{eqnarray}  Then, we can use the identity \begin{equation} \int_{-\infty}^{\infty} du \sqrt{\frac{a}{\pi}} e^{-a(x-u)^2}\sqrt{\frac{b}{\pi}} e^{-b(u -y)^2} = \sqrt{\frac{ab}{\pi(a+b)}} e^{-\frac{ab}{a+b}(x-y)^2} \end{equation} to evaluate the sum to be \begin{equation} Z = \sqrt{\frac{1}{2\pi \beta}}. \end{equation}
58097
From any textbook on fourier analysis:   "It is easily shown that for $f$ and $g$, both $2 \pi$-periodic functions on $[-\pi,\pi]$, we have $$(f \ast g)(x) = \int_{-\pi}^{\pi}f(x-y)g(y)\;dy = \int_{-\pi}^{\pi}f(z)g(x-z)\;dz = (g \ast f)(x),$$ by using the substitution $z = x-y.$"   I don't doubt that this is true, but I cannot figure out what happened to the negative sign coming from $dy = -dz\;$ after the change of variable $z = x - y$.  In particular, after the change of variable $z = x-y,\;$ I am coming up with $ -\int_{-\pi}^{\pi}f(z)g(x-z)\;dz$. What am I missing here?