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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}
\end{displaymath} (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).
\end{displaymath} (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

$\displaystyle \mathop{\rm sinc}\nolimits (nx)$ $\textstyle \equiv$ $\displaystyle {\sin(nx)\over nx} = {1\over nx} {e^{inx}-e^{-inx}\over 2i}$  
  $\textstyle =$ $\displaystyle {1\over 2inx} [e^{itx}]^n_{-n} = {1\over 2n} \int^n_{-n} e^{ixt}\,dt.$ (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).
\end{displaymath} (4)

Definite integrals involving the sinc function include
$\displaystyle \int_0^\infty \mathop{\rm sinc}\nolimits (x)\,dx$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}\pi$ (5)
$\displaystyle \int_0^\infty \mathop{\rm sinc}\nolimits ^2(x)\,dx$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}\pi$ (6)
$\displaystyle \int_0^\infty \mathop{\rm sinc}\nolimits ^3(x)\,dx$ $\textstyle =$ $\displaystyle {\textstyle{3\over 8}}\pi$ (7)
$\displaystyle \int_0^\infty \mathop{\rm sinc}\nolimits ^4(x)\,dx$ $\textstyle =$ $\displaystyle {\textstyle{1\over 3}}\pi$ (8)
$\displaystyle \int_0^\infty \mathop{\rm sinc}\nolimits ^5(x)\,dx$ $\textstyle =$ $\displaystyle {\textstyle{115\over 384}}\pi.$ (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,
\end{displaymath} (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{},
\end{displaymath} (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

{\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}
\end{displaymath} (12)


\begin{figure}\begin{center}\BoxedEPSF{Sine_Integral.epsf scaled 1000}\end{center}\end{figure}

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,
\end{displaymath} (13)

where $\Im$ denotes the Imaginary Point. Now define
$\displaystyle I$ $\textstyle \equiv$ $\displaystyle \int_\gamma {e^{iz}\over z}\,dz$  
  $\textstyle =$ $\displaystyle \lim_{R_1\to 0} \int_\pi^0 {\mathop{\rm exp}\nolimits (iR_1e^{i\theta })\over R_1e^{i\theta }}\, i\theta R_1 e^{i\theta }\,d\theta$  
  $\textstyle \phantom{=}$ $\displaystyle \mathop{+} \lim_{R_1\to 0}\lim_{R_2\to\infty} \int_{R_1}^{R_2} {e^{iR}\over R}\,dR$  
  $\textstyle \phantom{=}$ $\displaystyle \mathop{+} \lim_{R_2\to\infty} \int_0^\pi {\mathop{\rm exp}\nolimits (iz)\over z}\,dx
+ \lim_{R_1\to 0} \int_{R_2}^{R_1} {e^{-iR}\over -R} \,(-dR),$  

where the second and fourth terms use the identities $e^{i0}=1$ and $e^{i\pi}=-1$. Simplifying,
$\displaystyle I$ $\textstyle =$ $\displaystyle \lim_{R_1\to 0} \int_\pi^0 \mathop{\rm exp}\nolimits (iR_1e^{i\theta})i\theta \,d\theta+\int_{0^+}^\infty {e^{iR}\over R}\,dR$  
  $\textstyle \phantom{=}$ $\displaystyle + \lim_{R_2\to \infty} \int_0^\pi {\mathop{\rm exp}\nolimits (iz)\over z}\,dz+ \int_\infty^{0^+} {e^{-iR}\over -R}\,(-dR)$  
  $\textstyle =$ $\displaystyle -\int_0^\pi i\theta \,d\theta +\int_{0^+}^\infty {e^{iR}\over R}\, dR
+ 0+\int^{0^-}_{-\infty} {e^{iR}\over R}\,dR,$  

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.
\end{displaymath} (16)

Rearranging gives
\int_{-\infty}^\infty {e^{iz}\over z}\,dz = i\pi,
\end{displaymath} (17)

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

The same result is arrived at using the method of Residues by noting
$\displaystyle I$ $\textstyle =$ $\displaystyle 0+{\textstyle{1\over 2}}2\pi i\,{\rm Res} [f(z)]_{z=0}$  
  $\textstyle =$ $\displaystyle i\pi\left[{(z-0) {e^{iz}\over z}}\right]_{z=0} = i\pi [e^{iz}]_{z=0}$  
  $\textstyle =$ $\displaystyle i\pi,$ (19)

\end{displaymath} (20)

Since the integrand is symmetric, we therefore have
\int_0^\infty {\sin x\over x}\,dx={\textstyle{1\over 2}}\pi,
\end{displaymath} (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.
\end{displaymath} (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


Kogan, S. ``A Note on Definite Integrals Involving Trigonometric Functions.''

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

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© 1996-9 Eric W. Weisstein