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For $f=1_{[-1,1]}$, part (b) gives $\hat{f}(u)=2\sin u/u$. The Plancherel theorem in the stated Fourier convention givesSince the left side is two,The integrand is even, so the sinc-squared integral is

Use the Fourier transform convention $\hat{f}(k)={\int }_{\mathbb{R}}f(t)e^{-ikt}dt$, with inverse factor $1/(2\pi )$. For functions in L1 space, their convolution is $h(t)={\int }_{\mathbb{R}}f(s)g(t-s)ds$. Fubini's theorem applies because the absolute double integral is $\Vert f{\Vert }_1\Vert g{\Vert }_1$. Substituting $r=t-s$ therefore proves the convolution theorem:For the specified rectangular pulse, the support condition is $|t|<1$. Its Fourier transform isThe convolution $f\ast f$ is the length of the overlap of $(-1,1)$ and $(t-1,t+1)$. Thus it is the triangular pulseThe transform's value at zero is four, given by its continuous extension. In this convention, the Parseval identity for square-integrable functions isFor the rectangular pulse the left side is two, giving the sinc-squared integralFor the triangular pulse, $\int |f\ast f{|}^{2}dt=2{\int }_0^2(2-t{)}^{2}dt=16/3$. Apply the same Parseval identity with transform $4{\sin }^{2}k/k^2$ to obtain the sinc fourth-power integralBoth integrands have removable singularities at zero and integrable tails.

The triangular function $f(x)=(1-|x|)$ on $|x|<1$, zero elsewhere, has Fourier transform $\hat{f}(k)=[\sin (k/2)/(k/2){]}^2$ in the angular-frequency convention. Its squared L2 norm is $2/3$. Parseval's identity therefore givesThis proves the displayed value. The removable value at zero is one, and the fourth-power decay bounds the tail. The sinc-squared integral follows similarly from a rectangular function rather than a triangle.