Sinc Function

A function also called the Sampling Function and defined by

$$\mathop{\rm sinc}\nolimits (x) \equiv\cases{ 1 & for $x=0$\cr {\sin x\over x} & otherwise,\cr}$$ (1)

where $\sin x$ is the Sine function. Let $\Pi(x)$ be the Rectangle Function, then the Fourier Transform of $\Pi(x)$ is the sinc function

$${\mathcal F}[\Pi(x)]=\mathop{\rm sinc}\nolimits (\pi k).$$ (2)

The sinc function therefore frequently arises in physical applications such as Fourier transform spectroscopy as the so-called Instrument Function, which gives the instrumental response to a Delta Function input. Removing the instrument functions from the final spectrum requires use of some sort of Deconvolution algorithm.

The sinc function can be written as a complex Integral by noting that

$$\mathop{\rm sinc}\nolimits (nx) \equiv {\sin(nx)\over nx} = {1\over nx} {e^{inx}-e^{-inx}\over 2i}$$

$$= {1\over 2inx} [e^{itx}]^n_{-n} = {1\over 2n} \int^n_{-n} e^{ixt}\,dt.$$ (3)

The sinc function can also be written as the Infinite Product

$${\sin x\over x}=\prod_{k=1}^\infty \cos\left({x\over 2^k}\right).$$ (4)

Definite integrals involving the sinc function include

$$\int_0^\infty \mathop{\rm sinc}\nolimits (x)\,dx = {\textstyle{1\over 2}}\pi$$ (5)

$$\int_0^\infty \mathop{\rm sinc}\nolimits ^2(x)\,dx = {\textstyle{1\over 2}}\pi$$ (6)

$$\int_0^\infty \mathop{\rm sinc}\nolimits ^3(x)\,dx = {\textstyle{3\over 8}}\pi$$ (7)

$$\int_0^\infty \mathop{\rm sinc}\nolimits ^4(x)\,dx = {\textstyle{1\over 3}}\pi$$ (8)

$$\int_0^\infty \mathop{\rm sinc}\nolimits ^5(x)\,dx = {\textstyle{115\over 384}}\pi.$$ (9)

These are all special cases of the amazing general result

$$\int_0^\infty {\sin^a x\over x^b}\,dx = {\pi^{1-c}(-1)^{\lef... ...\right\rfloor -c} (-1)^k{a\choose k}(a-2k)^{b-1}[\ln(a-2k)]^c,$$ (10)

where $a$ and $b$ are Positive integers such that $a\geq b>c$, $c\equiv a-b\ \left({{\rm mod\ } {2}}\right)$, $\left\lfloor{x}\right\rfloor$ is the Floor Function, and $0^0$ is taken to be equal to 1 (Kogan). This spectacular formula simplifies in the special case when $n$ is a Positive Even integer to

$$\int_0^\infty {\sin^{2n}x\over x^{2n}}\,dx={\pi\over 2(2n-1)!}\left\langle{2n-1\atop n-1}\right\rangle{},$$ (11)

where $\left\langle{n\atop k}\right\rangle{}$ is an Eulerian Number (Kogan). The solution of the integral can also be written in terms of the Recurrence Relation for the coefficients

$$c(a,b)=\cases{ {\pi\over 2^{a+1-b}}{a-1\choose {\textstyle{... ... (b-1)(b-2)} [(a-1)c(a-2,b-2)-a\cdot c(a,b-2)] & otherwise\cr}$$ (12)

(Zimmerman).

The half-infinite integral of $\mathop{\rm sinc}\nolimits (x)$ can be derived using Contour Integration. In the above figure, consider the path $\gamma\equiv \gamma_1+\gamma_{12}+\gamma_2+\gamma_{21}$. Now write $z=Re^{i\theta}$. On an arc, $dz=iRe^{i\theta }\,d\theta$ and on the x-Axis, $dz=e^{i\theta}\,dR$. Write

$$\int_{-\infty}^\infty {\sin x\over x}\,dx = \Im \int_\gamma {e^{iz}\over z}\,dx,$$ (13)

where $\Im$ denotes the Imaginary Point. Now define

$$I \equiv \int_\gamma {e^{iz}\over z}\,dz$$

$$= \lim_{R_1\to 0} \int_\pi^0 {\mathop{\rm exp}\nolimits (iR_1e^{i\theta })\over R_1e^{i\theta }}\, i\theta R_1 e^{i\theta }\,d\theta$$

$$\phantom{=} \mathop{+} \lim_{R_1\to 0}\lim_{R_2\to\infty} \int_{R_1}^{R_2} {e^{iR}\over R}\,dR$$

$$\phantom{=} \mathop{+} \lim_{R_2\to\infty} \int_0^\pi {\mathop{\rm exp}\nolimits (iz)\over z}\,dx + \lim_{R_1\to 0} \int_{R_2}^{R_1} {e^{-iR}\over -R} \,(-dR),$$ (14)

where the second and fourth terms use the identities $e^{i0}=1$ and $e^{i\pi}=-1$. Simplifying,

$$I = \lim_{R_1\to 0} \int_\pi^0 \mathop{\rm exp}\nolimits (iR_1e^{i\theta})i\theta \,d\theta+\int_{0^+}^\infty {e^{iR}\over R}\,dR$$

$$\phantom{=} + \lim_{R_2\to \infty} \int_0^\pi {\mathop{\rm exp}\nolimits (iz)\over z}\,dz+ \int_\infty^{0^+} {e^{-iR}\over -R}\,(-dR)$$

$$= -\int_0^\pi i\theta \,d\theta +\int_{0^+}^\infty {e^{iR}\over R}\, dR + 0+\int^{0^-}_{-\infty} {e^{iR}\over R}\,dR,$$ (15)

where the third term vanishes by Jordan's Lemma. Performing the integration of the first term and combining the others yield

$$I=-i\pi+\int_{-\infty}^\infty {e^{iz}\over z}\, dz = 0.$$ (16)

Rearranging gives

$$\int_{-\infty}^\infty {e^{iz}\over z}\,dz = i\pi,$$ (17)

so

$$\int_{-\infty}^\infty {\sin z\over z}\,dz = \pi.$$ (18)

The same result is arrived at using the method of Residues by noting

$$I = 0+{\textstyle{1\over 2}}2\pi i\,{\rm Res} [f(z)]_{z=0}$$

$$= i\pi\left[{(z-0) {e^{iz}\over z}}\right]_{z=0} = i\pi [e^{iz}]_{z=0}$$

$$= i\pi,$$ (19)

so

$$\Im(I)=\pi.$$ (20)

Since the integrand is symmetric, we therefore have

$$\int_0^\infty {\sin x\over x}\,dx={\textstyle{1\over 2}}\pi,$$ (21)

giving the Sine Integral evaluated at 0 as

$$\mathop{\rm si}(0)=-\int_0^\infty {\sin x\over x}\,dx=-{\textstyle{1\over 2}}\pi.$$ (22)

An interesting property of $\mathop{\rm sinc}\nolimits (x)$ is that the set of Local Extrema of $\mathop{\rm sinc}\nolimits (x)$ corresponds to its intersections with the Cosine function $\cos(x)$, as illustrated above.

See also Fourier Transform, Fourier Transform--Rectangle Function, Instrument Function, Jinc Function, Sine, Sine Integral

References

Kogan, S. ``A Note on Definite Integrals Involving Trigonometric Functions.'' http://www.mathsoft.com/asolve/constant/pi/sin/sin.html.

Morrison, K. E. ``Cosine Products, Fourier Transforms, and Random Sums.'' Amer. Math. Monthly 102, 716-724, 1995.