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? 
 A: 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, $$\int_{-\infty}^\infty \frac{\sin(x)}{x}\delta (x)dx = \frac{\sin(x)}{x}|_{x=0}=1.$$
A: You may be thinking Riemann integration. Lebesgue  integration, which is more powerful but is not taught in high school, can handle this situation.
Like any Riemann integral where the integration range is not a proper subset of the domain of the integrand, this integral, taken as Riemann, cannot be evaluated.
The function $f(x)$ has a removable singularity at $x=0$, as Hugo C Botós points out in his answer. That means the modified Riemann integral
$$ \int_{-\infty}^{\infty} g(x) \delta(x) dx =1$$
with
$$g(x)=
\begin{cases}
\frac{sin x}{x},&x\ne 0\\\\
1,&x=0
\end{cases}
$$
can be evaluated.
Lebesgue integration, so to speak, automates this, where a function's domain has gaps of measure 0 (like isolated values).
Different symbols for the types of integration would have removed this ambiguity, but apparently this is not very relevant.
