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greetings . i have two questions regarding the sinc function in the week limit , where it can be used as a nascent delta function.

the definition :

[tex] \lim_{\varepsilon \rightarrow 0}\frac{1}{\pi }\int_{-\infty}^{\infty}\frac{sin\left(\frac{x-x_{0}}{\varepsilon}\right) }{x-x_{0}} \phi(x)dx=\phi(x_{0}) [/tex] is said to be valid for any smooth function [itex]\phi(x)[/itex] with compact support . does that mean that the following is not valid :

[tex] \lim_{\varepsilon \rightarrow 0}\frac{1}{\pi }\int_{-\infty}^{\infty}\frac{sin\left(\frac{x-x_{0}}{\varepsilon}\right) }{x-x_{0}}xdx=x_{0} [/tex]

moreover . if we expand the sine function, we get :

[tex] \lim_{\varepsilon \rightarrow 0}\frac{1}{\pi }\int_{-\infty}^{\infty}\phi(x)\sum_{n=0}^{\infty}\frac{(-1)^n(x-x_{0})^{2n}}{(2n+1)!(\varepsilon)^{2n+1}}dx =\lim_{\varepsilon \rightarrow 0}\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!(\varepsilon )^{2n+1}} \int_{-\infty}^{\infty}(x-x_{0})^{2n}\phi(x) =\phi(x_{0})

[/tex]

is it legit to perform the integration term by term ??

the definition :

[tex] \lim_{\varepsilon \rightarrow 0}\frac{1}{\pi }\int_{-\infty}^{\infty}\frac{sin\left(\frac{x-x_{0}}{\varepsilon}\right) }{x-x_{0}} \phi(x)dx=\phi(x_{0}) [/tex] is said to be valid for any smooth function [itex]\phi(x)[/itex] with compact support . does that mean that the following is not valid :

[tex] \lim_{\varepsilon \rightarrow 0}\frac{1}{\pi }\int_{-\infty}^{\infty}\frac{sin\left(\frac{x-x_{0}}{\varepsilon}\right) }{x-x_{0}}xdx=x_{0} [/tex]

moreover . if we expand the sine function, we get :

[tex] \lim_{\varepsilon \rightarrow 0}\frac{1}{\pi }\int_{-\infty}^{\infty}\phi(x)\sum_{n=0}^{\infty}\frac{(-1)^n(x-x_{0})^{2n}}{(2n+1)!(\varepsilon)^{2n+1}}dx =\lim_{\varepsilon \rightarrow 0}\frac{1}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(2n+1)!(\varepsilon )^{2n+1}} \int_{-\infty}^{\infty}(x-x_{0})^{2n}\phi(x) =\phi(x_{0})

[/tex]

is it legit to perform the integration term by term ??

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