Help With Bromwich Inversion Formula Proof

Question

To prove(copied from handwritten notes so possibly wrong):

Bromwich Inversion Formula. Fix $x_0∈ℝ $. If $F$ is complex analytic on $\{z:\Re z > x_0\} $ and for every $x>x_0$, $y↦ F(x + iy )$ is an $L^1$ function, i.e. $$ ∫_{x-i∞ }^{x+i∞} |F(z)| \ \newcommand{\d}{\text{d}}\d z < \infty$$ then $f:[0,∞)→\mathbb{C}$ defined by $$f(t) :=\frac{1}{2π i} ∫_{x-i∞ }^{x+i∞} F(z)e^{zt} \ \d z $$ where $x>x_0$ is arbitrary, is a function such that $\mathcal{L}f = F$, where $\mathcal{L}$ is the Laplace transform, $\mathcal{L}f(z) := ∫_0^∞ f(t) e^{-zt} \ \d t$.

I know that $f$ defined above is analytic, since it is the uniform limit of analytic functions (as you lengthen the domain of integration to $x\pm i∞$). My issue is in justifying the penultimate step (coloured $\color{blue}{\text{blue}}$) in the following formal calculation: \begin{align} ∫_0^∞ f(t) e^{zt} \ \d t &= ∫_0^∞ \left[\frac{1}{2π i} ∫_{x-i∞ }^{x+i∞} F(\xi)e^{\xi t} \ \d \xi\right] e^{-zt} \ \d t \\ &= \frac{1}{2π i}∫_{\xi = x-i∞ }^{x+i∞}F(\xi)\underbrace{∫_{t=0}^∞ e^{\xi t} e^{-zt} \ \d t}_{= \frac{0-1}{\xi-z}} \ \d \xi \\ &= -\frac{1}{2π i}∫_{\xi = x-i∞ }^{x+i∞}\frac{F(\xi)}{\xi-z} \d \xi \\ &= \color{blue}{-\frac{1}{2π i}∫_{\gamma}\frac{F(\xi)}{\xi-z} \d \xi} &&(\gamma \text{ a closed curve around $z$}) \\ &= F(z) \end{align} I essentially can't decide on a useful contour of integration (but also it has been a while since I needed complex analysis). Any help would be greatly appreciated.

---

Edit: I'll take a semicircle around $z$. How would I bound the integral on the semicircular arc? For concreteness I define $$L_R = [x-iR, x+iR]\text{ and }C_R = \{x+Re^{iπ t} : t ∈[-1/2 ,1/2] \}$$ $∫_{L_R} \dots \d \xi → ∫_{x-i∞}^{x+i∞} \dots \d \xi$ so I need to show that
$$\int_{C_R} \frac{F(\xi)}{\xi-z} \d \xi \xrightarrow[R→\infty]{} 0 $$

My attempt at finishing this: \begin{align} \left|\int_{C_R} \frac{F(\xi)}{\xi-z} \d \xi \right| &\leq π R ∫_{-1/2}^{1/2} \left|\frac{F(x + Re^{iπ t})}{x+Re^{iπ t} - z}\right| \d t \leq C ∫_{-1/2}^{1/2} | F(x + Re^{iπ t}) | \ \d t \end{align}

and I am unsure how to use the '$L^1$ on vertical strips' assumption to conclude this is 0 in the limit.

Answer 1

Assume first that $t \gt 0$. To use Cauchy's theorem, close the integration contour with a circular arc to the left of radius $R$ as follows.

($D$ is opposite $C$ along the imaginary axis.) Now consider

$$\oint_{ABCDA} dz \, F(z) e^{z t} $$

where

$$\int_{AB} dz \, F(z) e^{z t}= \int_{x-i \sqrt{R^2-x^2}}^{x+i \sqrt{R^2-x^2}} ds \, F(s) e^{s t} $$

$$\int_{BC} dz \, F(z) e^{z t} = i R \int_{\pi/2-\arcsin{(x/R)}}^{\pi/2} d\theta \, e^{i \theta} \, F \left (R e^{i \theta} \right ) e^{R e^{i \theta} t} $$

$$\int_{CD} dz \, F(z) e^{z t} = i R \int_{\pi/2}^{3 \pi/2} d\theta \, e^{i \theta} \, F \left (R e^{i \theta} \right ) e^{R e^{i \theta} t} $$

$$\int_{DA} dz \, F(z) e^{z t} = i R \int_{3 \pi/2}^{3 \pi/2+\arcsin{(x/R)}} d\theta \, e^{i \theta} \, F \left (R e^{i \theta} \right ) e^{R e^{i \theta} t} $$

First, we can show that the integral over $CD$ vanishes. This is done by bounding the magnitude of the integral:

$$\begin{align}\left |\int_{CD} dz \, F(z) e^{z t} \right | &\le R \int_{\pi/2}^{3 \pi/2} d\theta \, \left | F \left (R e^{i \theta} \right ) \right | e^{R t \cos{\theta}}\\ &\le 2 R \operatorname*{max}_{\theta \in [0,\pi]} \left | F \left (i R e^{i \theta} \right ) \right | \int_0^{\pi/2} d\theta \, e^{-R t \sin{\theta}} \\ &\le 2 R \operatorname*{max}_{\theta \in [0,\pi]} \left | F \left (i R e^{i \theta} \right ) \right | \int_0^{\pi/2} d\theta \, e^{-2 R t \theta/\pi}\\ &\le \frac{\pi}{t} \operatorname*{max}_{\theta \in [0,\pi]} \left | F \left (i R e^{i \theta} \right ) \right |\end{align}$$

Thus, in order for the integral over $CD$ to vanish, $F$ must satisfy the following condition:

$$\lim_{R \to \infty} \operatorname*{max}_{\theta \in [0,\pi]} \left | F \left (i R e^{i \theta} \right ) \right | = 0$$

Now we consider the integral over $BC$. Note that here we do not have the benefit of a vanishing exponential as we did we $CD$. We will need another approach to show that this segment too vanishes as $R \to \infty$:

$$\begin{align}\left |\int_{BC} dz \, F(z) e^{z t} \right | &\le R \int_{\pi/2-\arcsin{(x/R)}}^{\pi/2} d\theta \, \left | F \left ( R e^{i \theta} \right ) \right | e^{R t \cos{\theta}} \\ &= R \int_0^{\arcsin{(x/R)}} d\theta \, \left | F \left (i R e^{-i \theta} \right ) \right | e^{R t \sin{\theta}} \\ &\le R \left |F(i R) \right | \int_0^{x/R +O[(x/R)^3]} d\theta \, e^{R t \theta} \left [1+O(\theta^3) \right ] \\ &= \frac{e^{x t}-1}{t}\left | F \left (i R \right ) \right | + O \left ( \frac1{R^2} \right ) \end{align}$$

To summarize, the integral over $BC$ vanishes as $R \to \infty$ under the same conditions as that for the integral over $CD$. The integral over $DA$ vanishes for the same reasons.

Thus, under the highlighted condition above, one may invoke Cauchy's theorem as you have done above and prove the inversion formula.

When $t \lt 0$, close to the right. In that case, there is only a single circular arc and the integral vanishes for the same reasons as discussed for the integral over $CD$.