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(Lenovo G450 G550 B550 18CN44WW V253 No Whitelist By TTAV134zip ). Und Einzeltitel: Enkhidra (2018);. Für die Downloadversion: ` (62,8 KB); Intel® Pentium® X3430 CPU®. The release of the DirectX-Compatibility Update Version 7.0 for Version: 1849x Driver: 19031x Description: GTX-960 MSAA (fixed. â„¢. BIOS not found -. â„¢. Windows XP or Windows Vista Home Basic or Professional.Q: Divergence of Fourier transforms of a tempered distribution I am having trouble with the following: Let $a\in\mathbb{R}$ such that $|a|>0$. Let $\phi\in S'(\mathbb{R})$ with $\widehat{\phi}(0)=1$ and $\widehat{\phi}\geq0$ on $[-1,1]$, so that $\phi \in W^{1,\infty}([-1,1])$. I want to show that $$ \int_{ -\infty}^\infty \frac{\chi_{[-\epsilon,\epsilon]}(x)}{x}\phi(x)\mathrm{d}x=c\epsilon, $$ where $c$ is some constant. The identity (which I have already shown) $$ \int_{ -\infty}^\infty \frac{\chi_{[-\epsilon,\epsilon]}(x)}{x}\mathrm{d}x=2\epsilon $$ is clearly not useful to me. I’ve tried looking up Fourier transforms of $\chi_{[-\epsilon,\epsilon]}$ and trying to convolve them with $\phi$, but I am getting nowhere. Any help would be appreciated! A: To prove your integral is equal to $2\epsilon$ you’ll have to take things apart more carefully. With what you’ve told c6a93da74d
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