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$$\delta(x) = \begin{cases}

\infty & \text{if } x = 0 \\

0 & \text{if } x \neq 0

\end{cases}$$

And, also,

$$\int_{-\infty}^{\infty}\delta(x)\,dx=1$$

Using Fourier Transformation, it can be shown that,

$$\delta(x)=\lim_{\Omega \rightarrow \infty}\frac{\sin{(\Omega x)}}{\pi x}$$

Let's look at the graph of ##y=\frac{\sin{(\Omega x)}}{\pi x}##, for large ##\Omega##.

So, from this graph, it seems,

$$\lim_{\Omega \rightarrow \infty}\frac{\sin{(\Omega x)}}{\pi x}\neq 0 \text{ , for } x\neq 0 $$

But, this is actually the fourier transformation of the function,

$$f(t)=1=\exp{(i\cdot 0\cdot t)}$$

So, intution tells me that the frequency spectrum of ##f(t)## should have infinite amplitude at ##x=0## and ##0## amplitude otherwise, just like the Dirac-delta function.

Now, how can I show that,

$$\lim_{\Omega \rightarrow \infty}\frac{\sin{(\Omega x)}}{\pi x} = 0 \text{ when, } x\neq 0$$

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# I Dirac Delta using Fourier Transformation

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