Sinc Function

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The sinc function sinc(x), also called the "sampling function," is a function that arises frequently in signal processing and the theory of Fourier transforms. The full name of the function is "sine cardinal," but it is commonly referred to by its abbreviation, "sinc." There are two definitions in common use. The one adopted in this work defines

 sinc(x)={1   for x=0; (sinx)/x   otherwise,
(1)

where sinx is the sine function, plotted above.

This has the normalization

 int_(-infty)^inftysinc(x)dx=pi.
(2)

This function is implemented in the Wolfram Language as Sinc[x].

SincReImAbs
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When extended into the complex plane, sinc(z) is illustrated above.

SincCos

An interesting property of sinc(x) is that the set of local extrema of sinc(x) corresponds to its intersections with the cosine function cos(x), as illustrated above.

The derivative is given by

 (dsinc(z))/(dz)=(cosz)/z-(sinz)/(z^2)
(3)

and the indefinite integral by

 intsinc(z)dz=Si(z),
(4)

where Si(z) is the sine integral.

Woodward (1953), McNamee et al. (1971), and Bracewell (1999, p. 62) adopt the alternative definition

sinc_pi(x)=sinc(pix)
(5)
={1 for x=0; (sin(pix))/(pix) otherwise.
(6)

The latter definition is sometimes more convenient as a result of its simple normalization,

 int_(-infty)^inftysinc_pi(x)dx=1.
(7)

That variant also satisfies the sum

 sum_(k=-infty)^inftysinc(pik)=1.
(8)

In addition, the binomial coefficient satisfies

 (0; x)=sinc_(pi)(x),
(9)

which is essentially a restatement of the reflection relation

 Gamma(1+x)Gamma(1-x)=(pix)/(sin(pix))=1/(sinc_(pi)(x))
(10)

of the gamma function (M. Somos, pers. comm., Oct. 26, 2006.)

The sinc function is closely related to the spherical Bessel function of the first kind j_n(x) and, in particular,

 sinc(x)=j_0(x),
(11)

and is given in terms of the Meijer G-function as

 sinc(z)=(sqrt(pi))/2G_(02)^(10)(1/2z,1/2|-; 0,1/2).
(12)

Let Pi(x) be the rectangle function, then the Fourier transform of Pi(x) is the sinc function

 F_x[Pi(x)](k)=sinc(pik).
(13)

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, for x!=0,

sinc(nx)=(sin(nx))/(nx)
(14)
=1/(nx)(e^(inx)-e^(-inx))/(2i)
(15)
=1/(2inx)[e^(itx)]_(-n)^n
(16)
=1/(2n)int_(-n)^ne^(ixt)dt,
(17)

and that sinc(nx) and the integral both equal 1 for x=0.

The sinc function can also be written as the infinite product

 sinc(x)=product_(k=1)^inftycos(x/(2^k)),
(18)

a result discovered in 1593 by Francois Viète (Kac 1959, Morrison 1995) and sometimes known as Euler's formula (Prudnikov et al. 1986, p. 757; Gearhart and Shulz 1990). It is also given by

 sinc(x)=product_(k=1)^infty(1-(x^2)/(k^2pi^2))
(19)

(Gearhart and Shulz 1990) and

 sinc(x)=product_(k=1)^infty[1-4/3sin^2(x/(3^k))]
(20)

(Prudnikov et al. 1986, p. 757).

Another product is given by

product_(k=1)^(infty)sinc((2pi)/(2k+1))=pi/(2K)
(21)
=0.1805504...
(22)

(OEIS A118253; Prudnikov et al. 1986, p. 757), where K=8.700... is the constant from polygon circumscribing.

Sums of powers of sinc(k) over the positive integers include

sum_(k=1)^(infty)sinc(k)=-1/2+1/2pi
(23)
sum_(k=1)^(infty)sinc^2(k)=-1/2+1/2pi
(24)
sum_(k=1)^(infty)sinc^3(k)=-1/2+3/8pi
(25)
sum_(k=1)^(infty)sinc^4(k)=-1/2+1/3pi
(26)
sum_(k=1)^(infty)sinc^5(k)=-1/2+(115)/(384)pi
(27)
sum_(k=1)^(infty)sinc^6(k)=-1/2+(11)/(40)pi.
(28)

The remarkable fact that the sums of sinc(k) and sinc(k)^2 are equal appears to have first been published in Baillie (1978). Amazingly, the pattern of these sums being equal to -1/2 plus a rational multiple of pi breaks down for the power n=7, where the sum equals

 sum_(k=1)^inftysinc^7(k)=-1/2+P(x),
(29)

where

 P(x)=1/(46080)pi(129423-201684pi+144060pi^2-54880pi^3+11760pi^4-1344pi^5+64pi^6).
(30)

The sinc function satisfies the identity

 int_(-infty)^inftysinc[pi(x-y)]sinc(piy)dy=sinc(pix).
(31)

Definite integrals involving the sinc function include

int_0^inftysinc(x)dx=1/2pi
(32)
int_0^inftysinc^2(x)dx=1/2pi
(33)
int_0^inftysinc^3(x)dx=3/8pi
(34)
int_0^inftysinc^4(x)dx=1/3pi
(35)
int_0^inftysinc^5(x)dx=(115)/(384)pi.
(36)

After dividing out the constant factor of pi, the values are again 1/2, 1/2, 3/8, 1/3, 115/384, 11/40, 5887/23040, 151/630, 259723/1146880, ... (OEIS A049330 and A049331; Grimsey 1945, Medhurst and Roberts 1965). These are all special cases of the amazing general result

 int_0^infty(sin^ax)/(x^b)dx=(pi^(1-c)(-1)^(|_(a-b)/2_|))/(2^(a-c)(b-1)!)sum_(k=0)^(|_a/2_|-c)(-1)^k(a; k)(a-2k)^(b-1)[ln(a-2k)]^c,
(37)

where a and b are positive integers such that a>=b>c, c=a-b (mod 2), |_x_| 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)/(x^(2n))dx=pi/(2(2n-1)!)<2n-1; n-1>,
(38)

where <n; k> 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)={pi/(2^(a+1-b))(a-1; 1/2(a-1))   for b=1 or b=2; (a[(a-1)c(a-2,b-2)-a·c(a,b-2)])/((b-1)(b-2))   otherwise.
(39)
SincContour

The half-infinite integral of sinc(x) can be derived using contour integration. In the above figure, consider the path gamma=gamma_1+gamma_(12)+gamma_2+gamma_(21). Now write z=Re^(itheta). On an arc, dz=iRe^(itheta)dtheta and on the x-axis, dz=e^(itheta)dR. Write

 int_(-infty)^inftysinc(x)dx=Iint_gamma(e^(iz))/zdz,
(40)

where I denotes the imaginary part. Now define

I=int_(-infty)^infty(e^(iz))/zdz
(41)
=lim_(R_1->0)int_pi^0(exp(iR_1e^(itheta)))/(R_1e^(itheta))iR_1e^(itheta)dtheta+lim_(R_1->0)lim_(R_2->infty)int_(R_1)^(R_2)(e^(iR))/RdR+lim_(R_2->infty)int_0^pi(exp(iz))/zdz+lim_(R_1->0; R_2->infty)int_(R_2)^(R_1)(e^(-iR))/(-R)(-dR),
(42)

where the second and fourth terms use the identities e^(i0)=1 and e^(ipi)=-1. Simplifying,

I=lim_(R_1->0)int_pi^0exp(iR_1e^(itheta))idtheta+int_(0^+)^infty(e^(iR))/RdR+lim_(R_2->infty)int_0^pi(exp(iz))/zdz+int_infty^(0^+)(e^(-iR))/(-R)(-dR)
(43)
=-int_0^piidtheta+int_(0^+)^infty(e^(iR))/RdR+0+int_(-infty)^(0^-)(e^(iR))/RdR,
(44)

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

 I=-ipi+int_(-infty)^infty(e^(iz))/zdz=0.
(45)

Rearranging gives

 int_(-infty)^infty(e^(iz))/zdz=ipi,
(46)

so

 int_(-infty)^infty(sinz)/zdz=pi.
(47)

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

I=0+1/22piiRes_(z=0)f(z)
(48)
=ipi(z-0)(e^(iz))/z|_(z=0)
(49)
=ipi[e^(iz)]_(z=0)
(50)
=ipi,
(51)

so

 I(I)=pi.
(52)

Since the integrand is symmetric, we therefore have

 int_0^infty(sinx)/xdx=1/2pi,
(53)

giving the sine integral evaluated at 0 as

 si(0)=-int_0^infty(sinx)/xdx=-1/2pi.
(54)

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