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, $$\int_{-\infty}^\infty \frac{\sin(x)}{x}\delta (x)dx = \frac{\sin(x)}{x}|_{x=0}=1.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.