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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.

For the chosen Fourier transform convention, Parseval's identity is $\int |f{|}^{2}dx=(2\pi {)}^{-1}\int |\tilde{f}{|}^{2}dk$. For smooth rapidly decaying functions its bilinear form follows by inserting the inverse transform of $g$ into $\int f\overline{g}dx$ and interchanging the integrals:Take $g=f$. Approximation by such functions extends the identity to L2 space, which includes the compact triangular function here.

For that triangle, $\int |f{|}^{2}dx=2{\int }_0^1(1-x{)}^{2}dx=2/3$, while $\tilde{f}(k)=[\sin (k/2)/(k/2){]}^2$. ThusgivingThe continuous limiting integrand at zero is one. This is the sinc fourth-power integral.