Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 1 / 8E (Further Complex Methods)

Use the one-sided Laplace transform $F(s)={\int }_0^{\infty }{e}^{-st}f(t)dt$. Since $f=0$ on $[0,1)$, substitution gives $\mathcal{L}[f(t+1)](s)=e^{s}F(s)$. ThusIts poles are $s=0$ and $s_n=-\log 2+2\pi in$, with residues of $e^{st}F(s)$ respectively two and $2e^{s_{n}t}/s_n$. For $t>1$ close the Bromwich inversion formula to the left, using horizontal heights avoiding those poles. The left-side integrand decays like $e^{st}/s$, and on the right-side horizontal ends like $e^{s(t-1)}/s$, so the added integrals vanish with the usual symmetric limiting contour. The residue theorem giveswhere the series is summed symmetrically and at jumps has the midpoint value.

For $t<1$ the vanishing can be checked directly in the inversion integral, rather than inferred from the recurrence. On any vertical line $\Re s=c>0$ expand the denominator geometrically, uniformly there:Each inversion term is $2^{1-j}(2\pi i{)}^{-1}{\int }_{c-i\infty }^{c+i\infty }{e}^{s(t-j-1)}ds/s$. If $t<1$, close this contour to the right: the exponential decays, and the pole at zero is outside, so every term is zero. Equivalently the geometric expansion bounds the full closing contour by the same decaying factors. Thus $f(t)=0$ for $t<1$.

The same integral gives $f(t)=2{\sum }_{j\ge 0}{2}^{-j}H(t-j-1)$ elsewhere. For $m<t<m+1$ with integer $m\ge 1$, this is $4(1-2^{-m})$, which directly verifies the difference equation. Neither a Laplace transform nor a Fourier series determines values at isolated jumps; the printed series fixes the standard convention $H(0)=1/2$, including $f(1)=1$.