# Tag Info

3

Actually this is an easy one, since $\sin(x) = \dfrac{-i}{2} (e^{ix} - e^{-ix})$, $$e^x \sin(x) = \dfrac{-i}{2} (e^{(1+i) x} - e^{(1-i) x}) = \dfrac{-i}{2} \sum_{n=0}^\infty \dfrac{(1+i)^n - (1-i)^n)}{n!} x^n$$ But in general, $$\left(\sum_{n=0}^\infty a_n x^n\right)\left(\sum_{n=0}^\infty b_n x^n\right) = \sum_{n=0}^\infty \sum_{j=0}^n a_j b_{n-j} x^n$$ ...

2

I don't think going from step (2) to step (3) is obvious. You can do it with a trick. I'm renaming the constants so that they're not confused with yours. For $\alpha > 0$ and $\beta > 0$, define: $$F(\beta)=\int_{0}^{\infty}e^{-\alpha x^{2}}\cos(\beta x)\,dx.$$ Notice that $$F'(\beta)=-\int_{0}^{\infty}e^{-\alpha x^{2}}x\sin(\beta x)\,dx ... 2 Yes, it is true. Define a (complex-valued) measure on \mathbb{R} by$$\mu(B) := \sum_{k \in \mathbb{Z}} c_k \delta_k(B), \qquad B \in \mathcal{B}(\mathbb{R}). \tag{1}$$Since \sum_{k \in \mathbb{Z}} |c_k|<\infty, \mu is a finite measure on \mathbb{R}. From (1) we see that the formula$$\int f(x)\left( \sum_{k \in \mathbb{Z}} c_k \delta_k(dx) ...

1

We will answer negatively to this question. Note that the elements in $A(\Bbb{R})$ are all uniformly continuous functions. So, to find a counter example, we only need to consider some discontinuous function $f$. Let $f(x)={\bf 1}_{[-1,1]}(x)$ then it is easy to see that $f\in X_p$ for every $p\in(1,+\infty]$ but $f\notin A(R)$. So $X_p\not\subset ... 1 Check out Wolfram|Alpha with FourierParameters to specify your version of the Fourier Transform. In short: your result is correct. 1 The addition of$i\eta$helps because what you end up with at that point is an exponential of the form$e^{i(\omega+i\eta-\xi_k)t} = e^{i(\omega-\xi_k)t}e^{-\eta t}$. Notice then that you have exponential decay due to$\eta$which allows integrals to converge nicely. Otherwise your integral, as it stands, is not convergent in the usual way (since$e^{it}$... 1 The problem here is that $$\int_0^\infty e^{i\alpha t}$$ does not converge, because$e^{i\alpha t} = \cos \alpha t + i\sin \alpha t$doesn't go to zero for$t \to \infty$. But if you add an additional exponential factor$e^{-\eta t}$, you get $$\int_0^\infty e^{-\eta t} e^{i\alpha t} \,dt = \int_0^\infty e^{i(\alpha +i\eta)t} \,dt$$ which does ... 1 The absolute value (modulus) of any complex exponential is$1$. The reason is that$|z| = \sqrt{z\bar{z}}$. If$z = e^{ikx_0}$, then$\bar{z} = e^{-ikx_0}$and so$z\bar{z} = e^{-ikx_0}e^{ikx_0} = e^{-ikx_0+ikx_0} = e^0 = 1$. These are actually the same.$\delta(x-x_0) = \delta(x_0-x)$and$\delta(-x-x_0) = \delta(x+x_0)\$. You can check these by integrating ...

1

If you expand this integral you should get: $$\int_0^\pi \cos(nt)\sin(t)dt = \frac{1}{4i}\int_0^\pi \left(e^{int}+e^{-int}\right)\left(e^{it}-e^{-it}\right)dt$$ $$= \frac{1}{4i}\int_0^\pi e^{i(n+1)t} - e^{i(n-1)t} + e^{-i(n-1)t} - e^{-i(n+1)t}dt$$ Note that this is exactly the sum of sines: $$= i\int_0^\pi \sin((n+1)t)-\sin((n-1)t)dt$$ Now perform the ...

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