# Integration of the cardinal sine

It is said that the integral of the cardinal sine is $$1$$. How do I integrate the cardinal sine?

$$\int_{-\infty}^{\infty} \frac{\sin(a)}{a} \, {\rm d} a$$

• Do you know Fourier Transform theory? Aug 9, 2014 at 8:12
• yea I do. This is from one of the properties of FT. Just needed to know how to do the integration Aug 9, 2014 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 $\int_\mathbb{R} \text{sinc}(x)dx = \left. \mathcal{F}(\text{sinc}(x))(f) \right|_{f=0} = 1$ Aug 9, 2014 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.
$4)$ $\arctan(\infty)=\frac{\pi}{2}$

• Mind explaining your use of Fubini theorem too? Sep 20, 2014 at 16:18
• Could you explain your argument for (2)? Why can you reverse the order of integration of those Riemann integrals? Feb 24, 2018 at 16:09
• This answer is a little glib, and Step 2) is not valid as such without some justification. In particular, the integral is improper both for Riemann & Lebesgue integration. Oct 26, 2019 at 20:32
• For those interested, the following suggests a more rigorous approach: math.stackexchange.com/q/404549/27978 Oct 26, 2019 at 21:11

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:

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

Note:

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

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:

$$I'(t)=-\frac{1}{t^2+1}$$

Hence,

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

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

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

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

Let $t \to 0^+$:

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

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

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.

• 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. Aug 9, 2014 at 9:42
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". $$\int^\infty_0e^{-xt}\sin{ax}\ dx =\int^\infty_0e^{-xt}\cdot\frac{e^{iax}-e^{-iax}}{2}\ dx =\int^\infty_0 e^{\alpha x}-e^{\alpha^* x}\ dx$$