Sinc-squared function

This article is about a particular function from a subset of the real numbers to the real numbers. Information about the function, including its domain, range, and key data relating to graphing, differentiation, and integration, is presented in the article.
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For functions involving angles (trigonometric functions, inverse trigonometric functions, etc.) we follow the convention that all angles are measured in radians. Thus, for instance, the angle of

$90{}^{\circ }$

is measured as

$\pi /2$

.

Definition

This function is defined as the composite of the square function and the sinc function. Explicitly, it is given as:

$x\mapsto (\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}x{)}^2$

Alternatively, it is given as:

$\stackrel{\mathrm{i}}{\mathrm{s}}x:=\{\begin{array}{cc}1,& x=0\\ ,& x\ne 0\\ \end{array}$

Graph

Here is a graph on the interval $[-3\pi ,3\pi ]$:

The graph is a little unclear, here is an alternative version where different scalings are used for the $x$-axis and $y$-axis:

Integration

First antiderivative

Denote by $\mathrm{S}\mathrm{i}(x)$ the function $\underset{0}{\overset{x}{\int }}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}tdt=\underset{0}{\overset{x}{\int }}(\sin t)/tdt$. We integrate $\stackrel{\mathrm{i}}{\mathrm{s}}$ and obtain the following answer in terms of $\mathrm{S}\mathrm{i}$:

$\underset{0}{\overset{x}{\int }}\stackrel{\mathrm{i}}{\mathrm{s}}tdt=\mathrm{S}\mathrm{i}(2x)-\frac{{\sin }^{2}x}{x}$

For the indefinite integral, we can put a $+C$ at the end..

We do this using integration by parts:

$\underset{0}{\overset{x}{\int }}\stackrel{\mathrm{i}}{\mathrm{s}}tdt=\underset{0}{\overset{x}{\int }}\frac{{\sin }^{2}t}{t^2}dt$

Take $1/t^{2}dt$ as the part to integrate. We get:

${[{\sin }^{2}t(\frac{-1}{t})]}_0^x-\underset{0}{\overset{x}{\int }}\sin (2t)\frac{-1}{t}dt$

This becomes:

$\frac{-{\sin }^{2}x}{x}+\underset{t\to 0}{lim}\frac{{\sin }^{2}t}{t}+\underset{0}{\overset{x}{\int }}\frac{\sin (2t)}{t}dt$

The limit expression is zero because ${\sin }^2$ has a zero of order 2 at zero. For the integration expression, set $u=2t$ and get $\underset{0}{\overset{2x}{\int }}\frac{\sin (u)}{u}du=\mathrm{S}\mathrm{i}(2x)$. Plugging back in, we get the desired answer.

Improper definite integral

We know that:

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{c}tdt=\underset{x\to \mathrm{\infty }}{lim}\mathrm{S}\mathrm{i}(x)=\frac{\pi }{2}$

Using this, we obtain that:

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\stackrel{\mathrm{i}}{\mathrm{s}}tdt=\underset{x\to \mathrm{\infty }}{lim}\mathrm{S}\mathrm{i}(2x)-\underset{x\to \mathrm{\infty }}{lim}\frac{{\sin }^{2}x}{x}=\frac{\pi }{2}-0=\frac{\pi }{2}$

Similarly:

$\underset{-\mathrm{\infty }}{\overset{0}{\int }}\stackrel{\mathrm{i}}{\mathrm{s}}tdt=\frac{\pi }{2}$

Overall:

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\stackrel{\mathrm{i}}{\mathrm{s}}tdt=\pi$

Note that the improper integral value is the same for the sinc function and its square. Roughly speaking, the sinc function is bigger than its square when both are positive, but the sinc function also takes negative values while its square does not, and so these differences balance out in the overall integration.