# Definite integral involving sinc functions

I am trying to evaluate the following integral: $$\int_{-\infty}^{+\infty}\mathrm{sinc}(8t)\mathrm{sinc^2(t)}\cos(8\pi t) \, \mathrm{d}t$$ assuming that the sinc function is defined as: $$\mathrm{sinc}(t) = \begin{cases} \frac{\sin(\pi t)}{\pi t} \ &x \neq 0\\ 1 &x = 0 \end{cases}$$ I have tried to solve the problem using Fourier transforms and their properties, so: $$\int_{-\infty}^{+\infty}\mathrm{sinc}(8t)\mathrm{sinc^2(t)}\cos(8\pi t) \, \mathrm{d}t = \int_{-\infty}^{+\infty}x(t) \, \mathrm{d}t = \mathcal{F}\{x(t)\}(0) = X(0)$$ Focusing on the integrand $$x$$, it can be written as: $$x(t) = \mathrm{sinc}(8t)\cos(8\pi t)\mathrm{sinc^2(t)} = \left[\mathrm{sinc}(8t)\frac{e^{j8\pi t} + e^{-j8\pi t}}{2} \right] \mathrm{sinc^2(t)} = x_1(t)x_2(t)$$ where: \begin{align*} x_1(t) &= \frac{e^{j8\pi t}}{2}\mathrm{sinc}(8t) + \frac{e^{-j8\pi t}}{2}\mathrm{sinc}(8t) \\ x_2(t) &= \mathrm{sinc}^2(t) \end{align*} Using the convolution property of the Fourier transform and writing the transforms with capital letters, it is: $$X(f) = X_1(f) * X_2(f)$$ The transform of $$x_1$$ is given by: $$X_1(f) = \frac{1}{2}\left[\frac{1}{8}\mathrm{rect}_8(f-4) + \frac{1}{8}\mathrm{rect}_8(f+4) \right] = \frac{1}{16}\left[ \mathrm{rect}_8(f-4) + \mathrm{rect}_8(f+4) \right]$$ The transform of $$x_2$$ is given by: $$X_2(f) = \mathrm{tri}_1(f)$$ The transform of $$x$$ can be written as: $$X(f) = \int_{-\infty}^{+\infty} X_1(\xi) X_2(f-\xi) \, \mathrm{d}\xi$$ which, evaluated in $$f = 0$$, gives: $$X(0) = \int_{-\infty}^{+\infty} X_1(\xi) X_2(-\xi) \, \mathrm{d}\xi = \int_{-\infty}^{+\infty} \frac{1}{16}\left[ \mathrm{rect}_8(\xi-4) + \mathrm{rect}_8(\xi+4) \right] \mathrm{tri}_1(-\xi) \, \mathrm{d}\xi$$ Since the tri function is even, the last equation can be rewritten as: $$X(0) = \frac{1}{16}\int_{-\infty}^{+\infty} \left[ \mathrm{rect}_8(\xi-4) + \mathrm{rect}_8(\xi+4) \right] \mathrm{tri}_1(\xi) \, \mathrm{d}\xi$$ In order to evaluate the integral, I have plotted the functions which define the integrand, keeping in mind that I have to use the following definition of rect: $$\mathrm{rect}_\Delta(t) = \begin{cases} 1 &\left|t\right| < \Delta/2 \\ 0 &\left|t\right| > \Delta/2 \end{cases}$$ Since this definition of rect doesn't include a value for $$t = \pm \Delta/2$$, I don't know how to evaluate the integral. Looking at the figure, where it looks like the two rects don't intersect properly, I'd say that the integrand is zero and so the answer should be zero, however I'm not sure about the rigor of this solution: can I really say that the intersection is null or should I see the sum of rects as a big rect whose base goes from -8 to 8?

I have tried to check my calculations solving the integral with Wolfram and I have noticed that I get two different answers ($$A$$) if I write the integrand as a product of sincs and cos ($$A = 0$$) or as a product of sins (sinc in explicit form) and cos ($$A = 1/16$$), probably due to the way sinc and rect are defined in Wolfram.

integral_(-∞)^∞ sinc(8 t) sinc(t)^2 cos(8 π t) dt = 0

integral_(-∞)^∞ (sin(8 π t) (sin(π t)/(π t))^2 cos(8 π t))/(8 π t) dt = 1/16 = 0.0625


Define $$I$$ as the integral of interest

\begin{align} I&=\int_{-\infty}^\infty \text{sinc}(8t)\text{sinc}^2(t)\cos(8\pi t)\,dt\\\\ &= \int_{-\infty}^\infty \text{sinc}(16t)\text{sinc}^2(t)\,dt \end{align}

Now, the Fourier transforms of $$f(t)=\text{sinc}(16 t)$$ and $$g(t)=\text{sinc}^2(t)$$ are given respectively by

\begin{align} \mathscr{F}\{f\}(\omega)&=\int_{-\infty}^\infty f(t)e^{i\omega t}\,dt\\\\ &=\frac1{32}\left(\text{sgn}(\omega+16\pi)-\text{sgn}(\omega-16\pi)\right) \end{align}

\begin{align} \mathscr{F}\{g\}(\omega)&=\int_{-\infty}^\infty g(t)e^{i\omega t}\,dt\\\\ &=\frac{2\pi -\omega}{2\pi}\left(\text{sgn}(\omega+2\pi)-\text{sgn}(\omega-2\pi)\right) \end{align}

Using the convolution theorem, we find that

\begin{align} I&=\frac1{2\pi}\int_{-\infty}^\infty \frac{ \text{sgn}(\omega +16\pi)-\text{sgn}(\omega-16\pi)}{32} \left( \frac{2\pi -\omega}{2\pi}\left(\text{sgn}(\omega+2\pi)-\text{sgn}(\omega-2\pi)\right)\right)\,d\omega\\\\ &=\frac1{64\pi}\int_{-2\pi}^{2\pi}\frac{2\pi -\omega}{2\pi}\,d\omega\\\\ &=\frac1{16} \end{align}

Note that WolframAlpha defines the sinc funtion as

$$\text{sinc}(x)=\frac{\sin(x)}{x}\ne \frac{\sin(\pi x)}{\pi x}$$

When accounting for this differnce in definition WA retunrs the correct answer (See Here) of $$1/16$$.

• Thank you for your answer, introducing the function $\mathrm{sinc}(16t)$ I get $X(0) = \frac{1}{16}\int_{-\infty}^{+\infty}\mathrm{rect}_{16}(\xi)\mathrm{tri}_1(\xi)\, \mathrm{d}\xi = \frac{1}{16}$, however I am not able to understand yet why I get $X(0) = 0$ in Wolfram if I write the sinc function in its implicit form and the same happens in MATLAB if I write the sinc in its explicit form.
– gwn
Jun 29 at 13:43
• You're welcome. My pleasure. Jun 29 at 13:48
• With WolframAlpha, I get (See Here) the correct answer. Note that the sinc funciotn is defined there as $\text{sinc(x)}=\frac{\sin(x)}{x}$, without the $\pi$ in the argument. Jun 29 at 13:52
• Hi Mark ! You type too fast for the old man. Jun 29 at 13:53
• @claudeleibovici Hi Claude. How are you my friend? Jun 29 at 16:46

Using the definition $$\text{sinc}(t)=\frac{\sin (\pi t)}{\pi t}$$ you face the problem of $$I=\int \frac{\sin ^2(\pi t) \sin (8 \pi t) \cos (8 \pi t)}{8 \pi ^3 t^3}\,dt=\int \frac{\sin ^2(\pi t) \sin (16 \pi t)}{16 \pi ^3 t^3}\,dt$$ Using multiple angle formulae, we have $$I=\frac 1{64\pi^2t}\Big[7 \cos (14 \pi t)-16 \cos (16 \pi t)+9 \cos (18 \pi t) \Big]+$$ $$\frac 1{128\pi^3t^2}\Big[\sin (14 \pi t)-2 \sin (16 \pi t)+\sin (18 \pi t)\Big]+$$ $$\frac 1{32\pi}\Big[49\, \text{Si}(14 \pi t)-128\, \text{Si}(16 \pi t)+81 \,\text{Si}(18 \pi t)\Big]$$ Only the last term remains when using the bounds and the result $$I=\int_{-\infty}^{+\infty} \frac{\sin ^2(\pi t) \sin (8 \pi t) \cos (8 \pi t)}{8 \pi ^3 t^3}\,dt=\frac 1{16}$$
Let $$a$$ be any positive value greater than or equal to $$1$$.
\begin{align} \int_{-\infty}^{\infty} \operatorname{sinc}(at) \operatorname{sinc}^{2}(t) \cos(a \pi t) \, \mathrm dt &= \int_{-\infty}^{\infty} \frac{\sin(a \pi t) \sin^{2}(\pi t)\cos(a \pi t)}{a \pi^{3} t^{3}} \, \mathrm dt \\ &= \int_{-\infty}^{\infty} \frac{\sin(2 a \pi t) \sin^{2}(\pi t)}{2a \pi^{3} t^{3}} \, \mathrm dt \\ &= \Im \, \operatorname{PV}\int_{-\infty}^{\infty}\frac{e^{i 2a \pi t} \sin^{2}(\pi t)}{2a \pi^{3} t^{3}} \, \mathrm dt \\ &= \Im \, i \pi \operatorname{Res} \left[\frac{e^{i 2a \pi z} \sin^{2}(\pi z)}{2a \pi^{3} z^{3}}, 0 \right] \tag{1}\\ &= \Im \, i \pi \lim_{z \to 0} \frac{e^{i 2a \pi z} }{2a \pi} \left(\frac{\sin(\pi z)}{\pi z} \right)^{2} \\ &= \frac{1}{2a} . \end{align}
$$(1)$$ If $$a \ge 1$$, then $$|e^{i 2a \pi z} \sin^{2}(\pi z)|$$ is small in the upper half-plane.