Calculate $\int_0^\infty\frac{\sin^2t}{t^2}dt$ using Fourier transform

Question

Calculate $\int_0^\infty\frac{\sin^2t}{t^2}dt$ using Fourier transform. Unfortunately, we have not learned Plancherel's Theorem. The only other hint that was given is $$\mathcal{F}[f(t)]=\sqrt{\frac{2}{\pi}}\frac{1-\cos(\lambda\pi)}{\lambda^2},$$ where $$f(t)=\begin{cases}t+\pi&t\in[-\pi,0]\\\pi-t&t\in(0,\pi]\end{cases},$$ and $f(t)=0$ otherwise. I know that $$\int_0^\infty\frac{\sin^2t}{t^2}dt=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\sin^2t}{t^2}dt,$$ and I know the answer is supposed to be $\pi/2$, but I am unsure of how to get started.

Answer 1

I solved it, thanks to the helpful comment from Frank Lu.

$$I=\int_0^\infty\frac{\sin^2t}{t^2}dt=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\sin^2t}{t^2}dt$$

$$f(t)=\begin{cases}t+\pi&t\in[-\pi,0]\\\pi-t&t\in(0,\pi]\\0&\text{otherwise}\end{cases}$$

\begin{equation*} \begin{aligned} \mathcal{F}^{-1}\bigg[\sqrt{\frac{2}{\pi}}\frac{(1-\cos\lambda\pi)}{\lambda^2}\bigg] & = \mathcal{F}^{-1}\bigg[\sqrt{\frac{2}{\pi}}\frac{(2\sin^2(\lambda\pi/2))}{\lambda^2}\bigg] \\ \frac{\pi}{2}\cdot\sqrt{\frac{2}{\pi}}\frac{(2\sin^2\lambda)}{\lambda^2} & = \mathcal{F}\bigg[f(\frac{2}{\pi}t)\bigg] \\ & = \frac{\sqrt{2\pi}\sin^2\lambda}{\lambda^2} \\ \Rightarrow \mathcal{F}^{-1}[f(2t/\pi)] & = \int_{-\infty}^\infty\frac{\sin^2\lambda}{\lambda^2}e^{i\lambda t}d\lambda \end{aligned} \end{equation*} At $t=0$, we have $$\int_{-\infty}^\infty\frac{\sin^2\lambda}{\lambda^2}d\lambda=\pi\Rightarrow\int_0^\infty\frac{\sin^2t}{t^2}dt=\frac{\pi}{2}$$