Question on the definition of dirac delta function

Question

I was reading Wikipedia on the Dirac delta function https://en.wikipedia.org/wiki/Dirac_delta_function and there is a part that says: $\delta$ function is expressed by $$ \delta(x) = 1/2 \pi \int_{- \infty}^{\infty} e^{i \alpha x} d \alpha. $$

I was wondering why this can be interpreted as a delta function, because to be a delta function $\delta(x)$ has to be $0$ when $x$ is not zero. However, given some fixed non-zero $x$, $$ \lim_{L \rightarrow \infty} 1/2 \pi \int_{- L}^{L} e^{i \alpha x} d \alpha $$ does not seem to be $0$.

I would greatly appreciate any explanation. Thank you very much!

Answer 1

Well it isn't zero because when you compute it, $$∫_{-L}^L e^{i\alpha x} \ \text{d}\alpha = \frac{e^{iLx} - e^{iLx}}{ix} = \frac{\sin(Lx)}{2x} = \frac{L}{2} \frac{\sin(Lx)}{Lx}$$ you see that there isn't a limit as $L→∞$. The expression is more subtle than that.

You may want to see the following related question, Dirac Delta function and normal distribution Roughly, the Gaussian distribution with variance tending to zero, has the 'pointwise' behaviour one asks of the 'dirac delta function'.

Also, the following PDF may be helpful. http://www.cse.yorku.ca/~kosta/CompVis_Notes/fourier_transform_Gaussian.pdf