Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 7 / 3 / b / Solution

Let $M=\sup |f|$ and $F_T(z)={\int }_0^{T}f(t)e^{-tz}dt$. Each $F_T$ is an entire function: on any compact set of $z$, differentiation under this finite-interval integral is justified by a bounded integrable majorant. We prove $\boxed{F_T(0)\to F(0)}$, which gives ordinary convergence of the improper integral, not just an Abel limit.

Fix $R>0$. Since $\Omega$ contains the imaginary segment from $-iR$ to $iR$ and is open, choose $0<\delta <R$ so small that its closed left strip of width $\delta$ belongs to $\Omega$. Let $C_{+}$ be the right semicircle $|z|=R$ oriented from $-iR$ to $iR$, and let $\Gamma$ be the left half-ellipse $z=\delta \cos \theta +iR\sin \theta$, $\pi /2\le \theta \le 3\pi /2$, oriented back to $-iR$. The enclosed right half-disk and thin left region lie in $\Omega$. The residue theorem, applied to the analytic difference with its simple pole at zero, givesFor $x=\mathrm{Re}z>0$,On the right circle, $|1+z^2/R^2|=2x/R$ and $|z|=R$. The exponential cancels the displayed damping, so the absolute integrand per unit arc length is at most $2M/R^2$. The right-arc contribution is therefore at most $M/R$ after the factor $1/(2\pi )$.

For the $F_T$ term on $\Gamma$, deform that left path to the left semicircle $C_{-}$ of radius $R$. The Cauchy integral theorem allows this because $F_T$ is an entire function and the region between the paths excludes the pole at zero. On $C_{-}$, $x<0$ andThe same factor $|1+z^2/R^2|=2|x|/R$ bounds this contribution by $M/R$. The arc endpoints can be omitted from the estimates, since they have zero arc length and the integrands are continuous there.

Finally, the $F$ term on the original $\Gamma$ tends to zero as $T\to \infty$. This compact path stays away from zero, so $F(z)(1+z^2/R^2)/z$ is bounded there. Its interior has $\mathrm{Re}z<0$, hence $e^{Tz}\to 0$, and $|e^{Tz}|\le 1$. Dominated convergence on the finite-length path gives the claim. Combining the three pieces yieldsSince $R$ is arbitrary, the difference tends to zero. This proves the Newman Tauberian theorem in the required setting; the proof crucially uses analytic continuation across every finite segment of the imaginary axis.