Stuck with integral $\int_{-\infty}^\infty \left( \frac{\sin(a t+b)}{at+b} \right)^2 \, dt$

Question

I am stuck with the following integral:

$$\int_{-\infty}^\infty \left( \frac{\sin(a t+b)}{at+b} \right)^2 \, dt$$

I would like to show that $\varphi(t)=\frac{\sin(at+b)}{at+b}$ belongs to $L^2(\mathbb{R})$ and/or $L^1(\mathbb{R})$, i.e. $\int_{-\infty}^\infty | \varphi |^2 \,dt < \infty$ and/or $\int_{-\infty}^\infty | \varphi | \,dt < \infty $.

So far, I know that it is $|\frac{\sin(at+b)}{at+b}| \leq |\frac{1}{at+b}|$, but as $\int_{-\infty}^\infty |\frac{1}{at+b}|^2 \, dt$ does not converge, I cannot be conclusive. Looking at the plot it can be stated that it converges and then $\varphi \in L^2(\mathbb{R})$.

Answer 1

One standard strategy is$$\int_{-\infty}^\infty\frac{\sin^2 y}{y^2}dy=\int_{-\infty}^\infty\frac{e^{2iy}+e^{-2iy}-2}{-4}\int_0^\infty ze^{-zy}dzdy=\int_{-\infty}^\infty\frac{\frac{z}{z-2i}+\frac{z}{z+2i}-2}{-4}dz.$$I'll leave the rest to you.

Answer 2

Under the obvious substitution, $$\int_{-\infty}^\infty\frac{\sin^2(at+b)}{(at+b)^2}\,dt =\frac1{|a|}\int_{-\infty}^\infty\frac{\sin^2 t}{t^2}\,dt.$$ The integrand is bounded near zero, and $O(t^{-2})$ at infinity, so the integral converges.

By one of those amazing coincidences, $$\int_{-\infty}^\infty\frac{\sin^2 t}{t^2}\,dt =\int_{-\infty}^\infty\frac{\sin t}{t}\,dt=\cdots$$ a very well-known integral.

Answer 3

Yet another strategy: once the original integral has been reduced to $\frac{2}{|a|}\int_{0}^{+\infty}\frac{\sin^2 x}{x^2}\,dx$, one may invoke $\mathcal{L}(\sin^2 x)(s)=\frac{2}{s(4+s^2)}$ and $\mathcal{L}^{-1}\left(\frac{1}{x^2}\right)(s)=s$ to further reduce it to $$ \frac{4}{|a|}\int_{0}^{+\infty}\frac{ds}{s^2+4} = \frac{\pi}{|a|}.$$

Answer 4

I finally came out with this solution:

I computed the Fourier transform of $\varphi(t)$, which is: $$\hat{\varphi}(w) = \sqrt{\frac{\pi}{2} } \frac{e^{iwb/a}}{a}\chi_{[-a,a]}(w)$$

Then I learned that it is bandlimited, and as $\varphi \in L^2(\mathbb{R})$, then $\varphi$ lives in the Paley-Wiener space $PW_{a}$. Therefore, as in $L^2[-a,a]$ the Fourier transform is a unitary operator, it holds that:

$$\langle x,y \rangle = \langle \mathcal{F}x, \mathcal{F}y \rangle$$ so, $$I = \int_{-\infty}^{+\infty}{\left( \frac{sen(at+b)}{at+b} \right)^2 dt}$$

and: $$I = \langle \varphi, \varphi \rangle_{L^2({\mathbb{R}})} = \langle \hat{\varphi}, \hat{\varphi}\rangle_{L^2[-a,a]} = \frac{\pi}{2} \frac{1}{a^2} ||\chi_{[-a,a]}||^2 = \frac{\pi}{2} \frac{2a}{a^2} = \frac{\pi}{a}$$