Using Feynman Integration trick we can find the value of this integral to be equal to $\pi$.
$$\int_{-\infty}^{\infty} \frac{\sin(x)}{x}=\pi$$
But if we try the below approach:
$$I(a)=\int_{-\infty}^{\infty} \frac{\sin(ax)}{x}$$
$$I'(a)=\int_{-\infty}^{\infty}\cos(ax) $$
The integral does not converge.
We know that $I(1)=\pi$ and $I(0)=0$. Why is the integral not converging?
I saw the above trick to integrate, in this problem:$$\int_{-\infty}^{\infty} \frac{\sin^{2}(x)}{x^2}$$
Different approach to Feynman Integration trick
integrationmathematical physics
Best Answer
As defined, $I(a)=\pi\operatorname{sgn}a$ so $I^\prime(a)=2\pi\delta(a)$. Indeed, the identity$$\int_{\Bbb R}\exp(ikx)dx=2\pi\delta(k)$$implies as much, using $\cos(ax)=\frac12\sum_\pm\exp(\pm ikx)$ plus $\delta(-k)=\delta(k)$. But the problem is with the sense in which the above display-line equation "converges". Let's start with something completely uncontroversial: for any $\epsilon>0$,$$\int_{-1/\epsilon}^{1/\epsilon}\exp(ikx)dx=\frac{2}{k}\sin\frac{k}{\epsilon}=\frac{2}{\epsilon}\operatorname{sinc}\frac{k}{\epsilon}.$$This is a nascent delta function, so the distributional $\epsilon\to0^+$ limit gives the above result in terms of the Dirac delta.