Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2008 / Paper 1 / 4

Use the angular-frequency Fourier transform conventionFor the piecewise smooth functions here, the inverse gives the function at continuity points and the mean of one-sided limits at a jump. Values assigned at individual points do not affect the transform.

The even extension is the triangular function $f(x)=(1-|x|)$ for $|x|<1$ and zero outside. Its Fourier transform is real and even:At zero the removable limiting value is one, also obtained directly by integrating the triangle. This is the Fourier transform of a triangular function.

For the odd extension, the cosine integral vanishes and the sine integral doubles. Integration by parts givesThe value at $k=0$ is zero by the odd function symmetry, or by the Maclaurin series of the sine. The transform is purely imaginary and odd.

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.