# Integral involving Dirac Delta function

I have a basic question about integral involving dirac delta function. In my Signals and Systems textbook, it says that $$\int_{-\infty}^{\infty} f(x) \delta(x)dx = f(0)$$

But what if $f(x)$ is not defined on $x=0$? For example,

$\displaystyle{f(x)=\frac{\sin x}{x}}$.

From Wolframalpha, it says $$\int_{-\infty}^{\infty} \frac{\sin x}{x} \delta(x) dx =1$$ How do I evaluate such integral?

Every time you have a continuous function $f: \mathbb R \to \mathbb R$ you can define the integral $$\int_{-\infty}^\infty f(x)\delta (x)dx := f(0).$$
In the case of $f(x) = \sin(x)/x$, we have a continuous function defined on $\mathbb R$ if we define $f(0)=1$ because $\lim_{x\to 0}\sin(x)/x=1$. This limit can be computed using Taylor series for example since $sin(x) = x-x^3/3!+\cdots$.
So, $$\int_{-\infty}^\infty \frac{\sin(x)}{x}\delta (x)dx = \frac{\sin(x)}{x}|_{x=0}=1.$$