Solve $\int_0^{+\infty} \Big(\frac{\sin t}{t}\Big)^2 \ dt$

Question

$$\int_0^{+\infty} \Big(\frac{\sin t}{t}\Big)^2 \ dt=\frac{1}{2} \int_{-\infty}^{+\infty} \Big(\frac{\sin t}{t}\Big)^2 \ dt=\frac{1}{2} \Big\rvert \Big\rvert \frac{\sin t}{t} \Big\rvert \Big\rvert_2^2 $$

Plancherel theorem:

$$2 \pi \Big\rvert \Big\rvert f \Big\rvert\Big\rvert_2^2=\Big\rvert \Big\rvert \mathscr{F}(f) \Big\rvert\Big\rvert_2^2$$

$$p_2(x)=\begin{cases} 1 \qquad x \in [-1,1] \\ 0 \qquad x \notin [-1,1] \end{cases}$$
$$\frac{1}{2} \Big\rvert \Big\rvert \frac{\sin t}{t} \Big\rvert \Big\rvert_2^2=\frac{1}{2} 2\pi \Big\rvert \Big\rvert \frac{1}{2} \ p_2 \Big\rvert \Big\rvert_2^2=\frac{\pi}{2} $$

Is there any other method to calculate this integral?

Thanks!

Answer 1

Application of the convolution theorem provides another way forward. Let $f(t)$ and $F(\omega)$ constitute the Fourier Transform pair as given by

$$\begin{align}f(x)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty F(\omega)e^{-i\omega t}\,d\omega\\\\ F(\omega)&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(t)e^{i\omega t}\,dt \end{align}$$

Note that we have the Fourier Transform pairs

$$\begin{align} f(x) &\leftrightarrow F(\omega)\\\\ f^2(x) &\leftrightarrow \frac{1}{\sqrt {2\pi}}F(\omega)*F(\omega)\\\\ \frac{\sin(x)}{x}&\leftrightarrow \sqrt{\frac{\pi}{2}}\text{rect}(\omega/2)\\\\ \frac{\sin^2(x)}{x^2}&\leftrightarrow \frac{1}{\sqrt{2\pi}}\left(\sqrt{\frac{\pi}{2}}\text{rect}(\omega/2)\right)*\left(\sqrt{\frac{\pi}{2}}\text{rect}(\omega/2)\right)=\sqrt{\frac\pi8}2 \text{tri}(\omega/2) \tag 1 \end{align}$$

where $\text{rect}(t)$ and $\text{tri}(t)$ are the Rectangle Function and Triangle Function, respectively.

Finally, note that from $(1)$ we find

$$\int_0^\infty \frac{\sin^2(x)}{x^2}\,dx=\frac12\times \sqrt{2\pi}\times \sqrt{\frac\pi8}\times 2=\pi/2$$

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NOTE:

In THIS ANSWER, I applied the same approach to evaluate the integral $$\int_0^\infty \frac{\sin^4(x)}{x^4}\,dx$$

Answer 2

What about the good old integration by parts? We have:

$$ \int_{-\infty}^{+\infty}\frac{\sin^2 x}{x^2}\,dx = \int_{-\infty}^{+\infty}\frac{2\sin(x)\cos(x)}{x} \stackrel{x\mapsto\frac{t}{2}}{=} \int_{-\infty}^{+\infty}\frac{\sin t}{t}\stackrel{\text{Dirichlet}}{=}\color{red}{\pi}.$$

As an alternative, we may use the Laplace transform.
Since $\mathcal{L}(\sin^2 x)=\frac{2}{s(4+s^2)}$ and $\mathcal{L}^{-1}\left(\frac{1}{x^2}\right)=s$, we have: $$ \int_{-\infty}^{+\infty}\frac{\sin^2 x}{x^2}\,dx = 2\int_{0}^{+\infty}\frac{2\,ds}{4+s^2} \stackrel{s\mapsto 2t}{=} 2\int_{0}^{+\infty}\frac{dt}{1+t^2}=\color{red}{\pi}.$$

Answer 3

Over the top approach:

Recoginze that

$$ \frac{\sin(t)}t=\frac{1}{2}\int_{-1}^1dxe^{i xt} $$

Therefore

$$ 2\int_{0}^{\infty}\frac{\sin(t)^2}{t^2}dt=\int_{-\infty}^{\infty}\frac{\sin(t)^2}{t^2}dt=\frac{1}{4}\int_{-1}^1dx'\int_{-1}^1dx\int_{-\infty}^{\infty} dte^{i (x+x')t}=\\\frac{\pi}{2}\int_{-1}^1dx'\int_{-1}^1dx\delta(x+x')=\frac{\pi}{2}\int_{-1}^1dx'\textbf{1}_{(1,-1)}=\pi $$

$\textbf{1}_A$ denotes the indicator function on the set $A$ and $\delta(u)$ is Dirac's delta distribution

QED

Answer 4

And yet another approach is to use contour integration. Note that using Cauchy's Integral Theorem, we can write

$$\begin{align} \int_0^\infty\frac{\sin^2(t)}{t^2}\,dt&=\frac14\int_{-\infty}^\infty \frac{1-\cos(2t)}{t^2}\,dt\\\\ &=\frac12\int_{-\infty}^\infty \frac{1-\cos(t)}{t^2}\,dt\\\\ &=\frac12 \lim_{L\to \infty,\epsilon\to 0}\text{Re}\left(\int_{-L}^{-\epsilon} \frac{1-e^{it}}{t^2}\,dt+\int_{\epsilon}^L \frac{1-e^{it}}{t^2}\,dt\right)\\\\ &=\frac12 \lim_{\epsilon\to 0}\int_0^\pi \frac{1-e^{i\epsilon^{i\phi}}}{\epsilon^2 e^{i2\phi}}\,i\epsilon e^{i\phi}\,d\phi-\frac12 \lim_{L\to \infty}\int_0^\pi \frac{1-e^{iL^{i\phi}}}{L^2 e^{i2\phi}}\,iL e^{i\phi}\,d\phi\\\\ &=\pi/2 \end{align}$$

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\large 1)$ \begin{align} \int_{0}^{\infty}{\sin^{2}\pars{t} \over t^{2}}\,\dd t & = {1 \over 2}\lim_{N \to \infty}\int_{-N\pi}^{N\pi} {\sin^{2}\pars{t} \over t^{2}}\,\dd t = {1 \over 2}\lim_{N \to \infty}\sum_{k = -N}^{N - 1}\int_{k\pi}^{\pars{k + 1}\pi}{\sin^{2}\pars{t} \over t^{2}}\,\dd t \\[5mm] & = {1 \over 2}\lim_{N \to \infty}\sum_{k = -N}^{N - 1}\int_{0}^{\pi}{\sin^{2}\pars{t} \over \pars{t + k\pi}^{2}}\,\dd t = {1 \over 2}\int_{0}^{\pi}\sin^{2}\pars{t} \sum_{k = -\infty}^{\infty}{1 \over \pars{t + k\pi}^{2}}\,\dd t \\[5mm] & = {1 \over 2}\int_{0}^{\pi}\sin^{2}\pars{t}\csc^{2}\pars{t}\,\dd t = \bbx{\ds{\pi \over 2}} \end{align}

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$\large 2)$ \begin{align} \int_{0}^{\infty}{\sin^{2}\pars{t} \over t^{2}}\,\dd t & = \int_{0}^{\infty}\sin^{2}\pars{t}\pars{\int_{0}^{\infty}x\expo{-tx}\,\dd x} \dd t = \int_{0}^{\infty}x\,\Re\int_{0}^{\infty} {1 - \expo{2t\,\ic} \over 2}\,\expo{-tx}\dd t\,\dd x \\[5mm] & = {1 \over 2}\int_{0}^{\infty}x\,{4/x \over 4 + x^{2}}\,\dd x = \int_{0}^{\infty}{\dd x \over x^{2} + 1} = \bbx{\ds{\pi \over 2}} \end{align}