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You've calculated the variable shift incorrectly: if $f(x)=\cos(ax)$, then $f(x-b) = \cos(ax-ab)$. So in fact $$\mathcal{F}( \cos(a(x-\pi/(2a))) ) = e^{-i(\pi/2)(\omega/a)} \sqrt{\frac{\pi}{2}} (\delta(\omega-a) + \delta(\omega+a)).$$ But by the defining property of $\delta$, the exponential on the first term is the same as $e^{-i\pi/2}$, that on the ...

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This is similar to looking at a symmetric or hermitian matrix $A$, and expanding a vector in the eigenvectors of this matrix $A$. The operator $$Lf=-f'',\;\;\; 0 \le x < \infty,\;\; f(0)=0,$$ is a self-adjoint operator on the domain consisted of all twice-absolutely continuous functions $f\in L^2[0,\infty)$ for which $f''\in L^... 2 I'll use$z=x+iy$instead of$\lambda$. One can show that$|F(z)|$is uniformly bounded on circles of radius$n + 1/2$,$n = 1, 2, 3, \ldots$, and$\lim_{n \to \infty } F((n+\frac 12)y) = 0$. (1) implies that$F$is constant (using the maximum modulus principle and Liouville's theorem). (2) then implies that$F$is identically zero. Without loss of ... 1 This seems to be exactly what Jeff Rauch does in his notes “Fourier Analysis from Complex Analysis”, pag.7: http://www.math.lsa.umich.edu/~rauch/555/fouriercomplex.pdf The only difference is that Rauch uses$\sin \pi \lambda$in place of your$e^{-i2\pi \lambda}-1$. To obtain the key uniform boundedness, he uses the “Cauchy inequalities”, but to be honest I ... 1 I consulted with my senior in university and got the answer. Let$ j = n $for simplicity and define $$G(x_n) = \exp(2\pi ibx_n - \pi r(x_n - a)^2).$$ Then,$ u \in L^2(\mathbb{R}^n) $satisfies the condition if and only if $$u(x) = v(x_1, \dots, x_{n - 1}) G(x_n) \quad (\text{for a.e. x \in \mathbb{R}^n }) \tag{ * }$$ for some$ v \in L^2(\mathbb{R}...

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Apply the Euler–Maclaurin summation formula. For $f\in C^1\big([1,N]\big)$, it reads $$\sum_{n=1}^N f(n)=\int_1^N f(x)\,dx+\frac{f(1)+f(N)}{2}+\int_1^N\left(\{x\}-\frac12\right)f'(x)\,dx,$$ where $\{x\}=x-\lfloor x\rfloor$ is the fractional part of $x$. For $f(x)=e^{ic\log x}$ with $c=2\pi ka$, we have $f'(x)=icf(x)/x$ and $$W_N:=\frac1N\sum_{n=1}^N e^{ic\... 1 Apply the Euler–Maclaurin summation formula. For f\in C^1\big([1,N]\big), it reads$$\sum_{n=1}^N f(n)=\int_1^N f(x)\,dx+\frac{f(1)+f(N)}{2}+\int_1^N\left(\{x\}-\frac12\right)f'(x)\,dx,$$where \{x\}=x-\lfloor x\rfloor is the fractional part of x. For f(x)=e^{2\pi ib\log x}, we have f'(x)=2\pi ib f(x)/x and$$W_N:=\frac1N\sum_{n=1}^N e^{2\pi ib\log ...

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Using the fact that $-|t-x| = x-t$ if $x≤t$ and $t-x$ when $x≥t$, you get \begin{align*} 2\,f(-t) &= e^{-t}\int_{-\infty}^t e^{x-x^2} \mathrm d x + e^t \int_t^{\infty} e^{-x-x^2} \mathrm d x \\ &= e^{-t}\int_{-\infty}^t e^{-(x-1/2)^2+1/4} \mathrm d x + e^t \int_t^{\infty} e^{-(x+1/2)^2+1/4} \mathrm d x \\ &= e^{1/4-t}\int_{-\infty}^{t-1/2} e^{-... 0 Let \{ P_n \}_{n=0}^{\infty} denote the Legendre polynomials, normalized so that \|P_n\|_{L^2[-1,1]}=1 for all n. Let f \in L^2[-1,1] and let \epsilon > 0 be given. Then there is a continuous function g \in C[-1,1] such that \|f-g\|_{L^2} < \epsilon/2. The Weierstrass approximation theorem gives the existence of a polynomial p such that ... 1\begin{align} \int_0^{\pi/2}\frac{|\sin(2nx)|}{x}dx &= \int_0^{n\pi}\frac{|\sin(x)|}{x}dx \tag{i} \\ &= \sum_{k=1}^n \int_{(k-1)\pi}^{k\pi} \frac{|\sin (x)|}{x}dx \tag{ii} \\ &= \sum_{k=1}^n\int_{0}^{\pi} \frac{|\sin (x + (k-1)\pi)|}{x + (k-1)\pi}dx \tag{iii} \\ &= \sum_{k=1}^n\int_{0}^{\pi} \frac{\sin (x)}{x + (k-1)\pi}dx \tag{iv} \end{...

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In order to use separation of variables, we can propose a solution of the form $$\omega(t,\eta)=𝑔(𝑡)ℎ(\eta)+\frac{𝐶𝑡}{A}$$ where $g,h$ are two unknown functions, that we want to find out. Then we have: $$\frac{\partial\omega}{\partial t}=𝑔'(𝑡)ℎ(\eta)+\frac{𝐶}{A} \quad and \quad \frac{\partial^2\omega}{\partial \eta^2}=𝑔(𝑡)ℎ''(\eta).$$ Finally ...

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Using the Fourier transform $$\hat{f}(t,\xi) = \int_{-\infty}^{\infty} f(t,\eta) \, e^{-i\xi\eta} \, d\eta.$$ we get $$A\partial_t\hat{f}(t,\xi) = -B\xi^2\,\hat{f}(t,\xi) + C\,2\pi\,\delta(\xi).$$ The homogeneous equation, $$A\partial_t\hat{f}(t,\xi) = -B\xi^2\,\hat{f}(t,\xi)$$ has solutions $$\hat{f}_h(t,\xi) = \hat{R}(\xi)\,e^{-B\xi^2t/A},$$ where $... 1 Requested Form of the Integral Let$T$be an orthogonal linear transformation where$T(\xi)=|\xi|(1,0,0,\dots,0). \begin{align} \int_{\mathbb{R}^d}\frac{\left|\,e^{i\langle\xi,y\rangle}+e^{-i\langle\xi,y\rangle}-2\,\right|^2}{|y|^{d+2}}\,\mathrm{d}y &=\int_{\mathbb{R}^d}\frac{\left|\,e^{i|\xi|y_1}+e^{-i|\xi|y_1}-2\,\right|^2}{|y|^{d+2}}\,\mathrm{d}y\... 1 Define F(\xi) = \int_{\mathbb{R}^d} \dfrac{|e^{i\langle \xi, y \rangle} + e^{- i\langle \xi, y \rangle} - 2|^2 }{|y|^{d+2}}dy. Show that F(\alpha \xi) = |\alpha|^{\alpha} F(\xi) for some \alpha \in \mathbb{R} (independent of \xi \in \mathbb{R}^{d} \setminus \{0\}), and then show that F(O(\xi)) = F(\xi) for each orthogonal transformation O. ... 0 This is getting here a little late (after the acceptance of a solution), but I thought I would expand on @uniquesolution's answer. To me, it seems that it is the continuity of \sigma_N f that is the issue. For that I will make use of the following result: Proposition: Let g \in L^1([-\pi,\pi]) and h \in C^0([-\pi,\pi]), then f \ast g \in C^0([-\pi,\... 1 Answering my own question here, so it may be false. Define,S_N(f)=\sum_{n=0}^{N}\langle f, \mathfrak{L_n}\rangle\mathfrak{L}_n(\theta)$$Where f is assumed to be Riemann integralable on [-1,1] and \mathfrak{L}_n(\theta) is defined in the question. Now consider the following inner product for m \leq N,$$\langle f-S_N(f),\mathfrak{L}_m \rangle=\... 0\hat{f}(z)$is$\frac{1}{2^n}$times the sum of$2^n$iid Bernoulli random variables taking on the values$1, -1$with probability$\frac{1}{2}$each, for every$z$(the signs$(-1)^{x \cdot z}$don't affect this!). So it has a binomial distribution: we have $$\mathbb{P} \left( \hat{f}(z) = \frac{2^n - 2k}{2^n} \right) = \frac{1}{2^{2^n}} {2^n \choose k}.$$ ... 0 It is actually simple: To the function$f(x)$defined on$[0,L)$, attach the function$g(x)=f(-x)$defined on$[-L,0)$. The resulting composite function$h(x)$is even ($h(x) = h(-x)$) on the interval$[-L,L)$, therefore its Fourier series has cosines only. This captures the DCT of$f(x)$. 1 Looking at the integral it is obvious that this is a convolution of :$f(p)$and$exp\left(\frac{-p^2}{2}\right)$Now take fourier transform on both sides to get: $$\mathcal{F}(exp\left(\frac{-x^2}{4}\right))=\mathcal{F}(exp\left(\frac{-p^2}{2}\right))\mathcal{F}(f(p))$$ $$\mathcal{F}(exp\left(\frac{-x^2}{4}\right))=\int _{-\infty }^{\infty }e^{-0.25x^2-2\pi ... 0 This is related to the following, where S_N^f(\theta) is the Fourier series truncated to the terms 1,\sin(n\theta),\cos(n\theta) for n=1,2,3,\cdots,N.$$ S_{N}^f(\theta)-L=\frac{1}{\pi}\int_0^{\pi}D_N(\theta')\left[\frac{f(\theta+\theta')+f(\theta-\theta')}{2}-L\right]d\theta. $$The Dirichlet-Dini condition is formulated to make sure that the ... 1 Let f be any function in L^{1} which is not in L^{2}. Take g(y)=f(-y) so that g is also in L^{1}. Then then convolution does not exist at x=0. 4 Reduce the problem by looking at the equation solved by$$ v(x,t)=u(x,t)-\left(1-\frac{x}{L}\right)T_f-\frac{x}{L}T_i $$This function v satisifes$$ v_t = v_{xx} \\ v(x=0,t > 0) = u(x=0,t)-T_f=0 \\ v(x=L,t > 0) = u(x=L,t)-T_i=0 \\ v(x,0)= f(x)-\left(1-\frac{x}{L}\right)T_f-\frac{x}{L}T_i $$With ... 3 That is not a function \mathbb{R} \to \mathbb{R} since \infty \not\in \mathbb{R}. It is a function \mathbb{R} \to \bar{\mathbb{R}} = \mathbb{R} \cup \{\infty\} though. 2 Let f(t) = \frac{1}{(1+t^2)(4+t^2)}. Its Fourier transform is$$ \hat{f}(x) = \int_{-\infty}^{\infty} \frac{e^{-itx}}{(1+t^2)(4+t^2)}dt = \int_{-\infty}^{\infty} \frac{1}{3}\left( \frac{1}{1+t^2} - \frac{1}{4+t^2} \right) e^{-itx} dt \\ = \frac{1}{12} \int_{-\infty}^{\infty} \left(2 \frac{2\cdot 1}{1+t^2} - \frac{2\cdot 2}{4+t^2} \right) e^{-itx} dt = \... 1 Fourier considered the integral representation to be the limit of a discrete series as the period tended to$\infty$. The Fourier cosine version is $$f(x) \sim \frac{2}{\pi}\int_{0}^{\infty}\left(\int_{0}^{\infty}f(y)\cos(sy)dy\right) \cos(sx) ds$$ This form is correct. It can be derived from the exponential form after extending$f$to ... 3 Suppose$x\ge0$. Let $$I(x)=\int_0^\infty \frac{t\sin(tx)}{(1+t^2)(4+t^2)}dt.$$ Then \begin{eqnarray} I'(x)&=&\int_0^\infty \frac{t^2\cos(tx)}{(1+t^2)(4+t^2)}dt\\ &=&\int_0^\infty \frac{\cos(tx)}{4+t^2}dt-\int_0^\infty \frac{\cos(tx)}{(1+t^2)(4+t^2)}dt\\ &=&\frac\pi4e^{-2x}-\int_0^\infty \frac{\cos(tx)}{(1+t^2)(4+t^2)}dt \end{... 1 It is not clear from your question what you are trying to do. I think you have three options: Use a change of variable such as$\theta = 2 \tan^{-1} x$which transforms$[0,\infty)$to$[0,\pi)$. Then for a function$f:[0,\infty)$you could obtain the cosine series for$f(\tan(\theta/2))=a_0/2 + \sum a_n \cos n \theta$. This results in a cumbersome ... 1 hint: With$f(x)=e^{-ax}$the definition of fourier transform shows $$\hat{f}(w)=\sqrt{\dfrac{2}{\pi}}\dfrac{a}{a^2+w^2}$$ then $$f(x)=\int_{-\infty}^{\infty}\hat{f}(w)e^{2\pi ixw}dw=a\sqrt{\dfrac{2}{\pi}}\int_{0}^{\infty}\dfrac{2i \sin(2\pi xw)}{a^2+w^2}dw=e^{-ax}$$ so from $$\dfrac{1}{(1+w^2)(4+w^2)}=\dfrac13\dfrac{1}{1+w^2}-\dfrac13\dfrac{1}{4+w^2}$$ you ... 1 First $$I(x):=\int_0^{\infty} \frac{t\sin(tx)}{(1+t^2)(4+t^2)}dt$$ then applying a Laplace transform we get $$\mathscr{L}_{x\to s}\{I(x)\}=\int_0^{\infty} e^{-sx}\int_0^{\infty} \frac{t\sin(tx)}{(1+t^2)(4+t^2)}dtdx$$ here by the$x\to s$I just denote which vairables I change to which, to avoid confusion. Now since all integrals converge we change order of ... 4 1st Solution. Following @Claude Leibovici's suggestion, let us utilize complex-analytic technique. First, symmetrize the integral to write $$I = \frac{1}{2}\int_{-\infty}^{\infty} \frac{t\sin(tx)}{(1+t^2)(4+t^2)}\,\mathrm{d}t = \frac{1}{2}\operatorname{Im}\biggl(\int_{-\infty}^{\infty}\frac{te^{itx}}{(1+t^2)(4+t^2)}\,\mathrm{d}t\biggr).$$ Now let$R > 2$... 3 Partial answer. $$I=\int_0^\infty \frac{t\sin{(tx)}}{(1+t^2)(4+t^2)}\,dt=\Im\left(\int_0^\infty \frac{t\,e^{itx}}{(1+t^2)(4+t^2)}\,dt \right)$$ $$\frac{t}{(1+t^2)(4+t^2)}=\frac{1}{6 (t+i)}-\frac{1}{6 (t-2 i)}-\frac{1}{6 (t+2 i)}+\frac{1}{6 (t-i)}$$ So, you basically face four integrals $$I_k=\int_0^\infty\frac {e^{itx}}{t+ki}$$ With simple change of ... 1 Perhaps item 1 should be written to imply that$f$is equal, almost everywhere, to some continuous function$g$(i.e., in the "equivalence class" (mod equality a.e.) of$f$, there is at least one continuous function.) 0 If the Fourier Series for$f$converges uniformly to$g$, then$gis continuous and \begin{align} \hat{g}(n)&=\frac{1}{2\pi}\int_{0}^{2\pi}g(x)e^{-inx}dx \\ &=\lim_{N\rightarrow\infty}\frac{1}{2\pi}\int_{0}^{2\pi}\sum_{k=-K}^{K}\hat{f}(k)e^{ikx}e^{-inx}dx \\ &= \lim_{N\rightarrow\infty}\sum_{k=-K}^{K}\hat{f}(k)\frac{1}{2\pi}\... 2 I find it suspicious thata_n=0$since the function$y-\dfrac{1}{2}$is not odd. According to your calculations$y-\dfrac{1}{2}$is a pure sinus wave. You should check the calculations of$a_n$when$n=8\$, the integral now involves $$\cos^2\left(\dfrac{16\pi t}{5}\right)$$ and is unlikely to vanish.

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