Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How to integrate sinc function. It is said that total area of a sinc function is 1. How do i Integrate sinc function. Or more simply $$\int_{-\infty}^{\infty} \frac{\sin(a)}{a}da $$

share|cite|improve this question
Do you know Fourier Transform theory? – the_candyman Aug 9 '14 at 8:12
yea I do. This is from one of the properties of FT. Just needed to know how to do the integration – Shivji Aug 9 '14 at 9:06
The integral of $\text{sinc}\ x=\dfrac{\sin\pi x}{\pi x}$ is 1. Indeed the Fourier Transform of $\text{sinc}(x)$ is the $\text{rect}(x)$ function, which is $1$ for $|x| < \frac{1}{2}$ and $0$ elsewhere. Recall that $\int_\mathbb{R} \text{sinc}(x)dx = \left. \mathcal{F}(\text{sinc}(x))(f) \right|_{f=0} = 1$ – the_candyman Aug 9 '14 at 19:05

Use the fact that $$\int^\infty_0e^{-xt}dt=\frac{1}{x}$$ Hence \begin{align} \int^\infty_{-\infty}\frac{\sin{x}}{x}dx \tag1 &=2\int^\infty_{0}\frac{\sin{x}}{x}dx\\ \tag2 &=2\int^\infty_0\int^\infty_0e^{-xt}\sin{x}dxdt\\ \tag3 &=2\int^\infty_0\frac{1}{1+t^2}dt\\ \tag4 &=\pi \end{align} Explanation:
$1)$Integrand is even
$2)$Reverse the order of integration
$3)$Recognise the laplace transform of $\sin{x}$, or integrate by parts.
$4)$ $\arctan(\infty)=\frac{\pi}{2}$

share|cite|improve this answer
Mind explaining your use of Fubini theorem too? – Troy Woo Sep 20 '14 at 16:18

If you are using the normalised $\mathrm{sinc}$ function, the area will be $1$ though if not, it is $\pi$. Proofs can be found here and here. Note that the second link still answers your question even though the integrand is squared.

Please consider googling your question before asking :)

share|cite|improve this answer
If the OP means the normalized sinc function, $\text{sinc}\ x=\dfrac{\sin\pi x}{\pi x}$ for $x\neq0$, it is true that its normalization is equal to $1$. At least, I was taught so when I took signal processing or Fourier analysis course. – Tunk-Fey Aug 9 '14 at 9:42
@Tunk-Fey Good point; I'll edit my answer. – Jam Aug 9 '14 at 9:49

Regarding the (great) answer of SuperAbound, Aug 9 '14: I think it is easier to solve integral (2) using Euler's identity and $\alpha = ia-t$ instead of "integrati[on] by parts". $$\int^\infty_0e^{-xt}\sin{ax}\ dx =\int^\infty_0e^{-xt}\cdot\frac{e^{iax}-e^{-iax}}{2}\ dx =\int^\infty_0 e^{\alpha x}-e^{\alpha^* x}\ dx$$

share|cite|improve this answer

A common method given:

Because the function $f(x)=\frac{\sin x}{x}$, where $f: \mathbb{R} - \{0\} \to \mathbb {R}$ is even we have:

$$\int_{-\infty}^{\infty} \frac{\sin x}{x} dx=2\int_{0}^{\infty} \frac{\sin x}{x} dx$$

Now let:

$$I(t)=\int_{0}^{\infty} \frac{\sin x}{x} e^{-tx} dx$$


$$\frac{\partial}{\partial t} \frac{\sin x}{x} e^{-tx}=\frac{\sin x}{x} e^{-tx}(-x)$$

So by differentiation under the integral sign we have:

$$I'(t)=-\int_{0}^{\infty} e^{-tx} \sin x dx$$

And through integration by parts twice we have:



$$I(t)=\int -\frac{1}{t^2+1} dt$$

$$I(t)=-\arctan (t) +c$$

But as $t \to \infty$, $I(t) \to 0$ hence:

$$I(t)=\frac{\pi}{2}-\arctan t$$

Let $t \to 0^+$:

$$\int_{0}^{\infty} \frac{\sin x}{x} dx=\frac{\pi}{2}$$

$$\int_{-\infty}^{\infty} \frac{\sin x}{x} dx=\pi$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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