Evaluate integral $ \int_{\mathbb{R} } \frac{\sin^4(\frac{t}{2}) }{t^2}$ Let $f(x)= (1-2|x|)\chi_{[-\frac{1}{2}, \frac{1}{2}]} (x)  $. Its Fourier transform is given by
$ \hat{f} (x) = \frac{8\sin^2(\frac{t}{4})}{t^2} $. Based on this, I need to evaluate the integral $ \int_{\mathbb{R} } \frac{\sin^4(\frac{t}{2}) }{t^2}$, but I don't know where to start.  I tried Parseval's identity - I wrote  $\frac{\sin^4(\frac{t}{2}) }{t^2} = \frac{\sin^2(\frac{t}{2}) }{t^2} \cdot \sin^2(\frac{t}{2})$, and then I would need to find the function whose Fourier transform is $\sin^2(t)$. I don't know if this is right way (seems too complicated), so is there any simple way to evaluate integral? I would be very grateful for help.
 A: I would guess it's a typo and there was supposed to be a $t^4$ in the denominator and you were supposed to use Plancherel's theorem. But you can do the original integral as follows:
Note that the integrand is even and that $\sin^2 (t/2) = {1 - \cos t \over 2}$. So your integral is
$$2\int_0^{\infty} {(1 - \cos t)^2 \over 4t^2}\,dt$$
You can integrate this by parts, integrating ${1 \over 4t^2}$ and differentiating $(1 - \cos t)^2$ (being careful that the improper integrals converge). Your integral becomes
$$2 \int_0^{\infty} {2(1 - \cos t)\sin t \over 4t}\,dt$$
Using $\sin 2t = 2\sin t \cos t$, this is
$$\int_0^{\infty} {\sin t \over t}\,dt -  {1 \over 2} \int_0^{\infty}{\sin 2t \over t}\,dt$$
Changing variables to $2t$ in the latter integral, this becomes
$$\int_0^{\infty} {\sin t \over t}\,dt -  {1 \over 2} \int_0^{\infty}{\sin t \over t}\,dt$$
$$= {\pi \over 2} - {\pi \over 4}$$
$$= {\pi \over 4}$$
A: First a simple substitution
$$\int_{\Bbb{R}}\frac{\sin^4(t/2)}{t^2}\mathrm{d}t=\int_{0}^\infty \frac{\sin^4(s)}{s^2}\mathrm{d}s$$
There is a known formula
$$\int_{0}^\infty\frac{\sin(x)^m}{x^n}\mathrm{d}x$$ $$=\frac{\pi^{1-p}(-1)^{\lfloor(m-n)/2\rfloor}}{2^{m-p}(n-1)!}\sum_{k=0}^{\lfloor m/2\rfloor-p}(-1)^k \mathrm{C}(m,k)(m-2k)^{n-1}\log(m-2k)^p$$
$p=\operatorname{mod}(m-n,2)$, which holds for $m,n\in\Bbb{N}$ and $m\geq n>p$. When using this expression we make the definition $0^0:=1$. In our case, $m=4,n=2,p=0$ so
$$\int_{0}^\infty \frac{\sin^4(s)}{s^2}\mathrm{d}s=\frac{-\pi}{2^4}\sum_{k=0}^{2}(-1)^k \mathrm{C}(4,k)(4-2k)$$
$$=\frac{-\pi}{2^4}\cdot(-4)=\frac{\pi}{4}$$
