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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 $$\int_{-\infty}^{\infty} \frac{\sin(a)}{a}da $$

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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.
$4)$ $\arctan(\infty)=\frac{\pi}{2}$

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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 :)

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2  
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". $$\int^\infty_0e^{-xt}\sin{ax}\ dx =\int^\infty_0e^{-xt}\cdot\frac{e^{iax}-e^{-iax}}{2}\ dx =\int^\infty_0 e^{\alpha x}-e^{\alpha^* x}\ dx$$

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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 $t \to 0^+$:

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

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

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