[proofplan] We evaluate the Bromwich inversion integral by closing the contour with a large semicircular arc and applying the residue theorem. For $t > 0$, we close to the left, where $|e^{pt}|$ decays along the arc; the ML inequality kills the arc contribution, and the residue theorem collects the singularities. For $t < 0$, we close to the right, where $|e^{pt}|$ decays instead; the closed contour encloses no singularities, yielding $f(t) = 0$. [/proofplan] [step:Close the Bromwich contour to the left and show the arc vanishes for $t > 0$] For $t > 0$, augment the vertical Bromwich line $\operatorname{Re} p = c$ with a semicircular arc $\Gamma_R$ of radius $R$ in the left half-plane, parametrised by $p = c + Re^{i\theta}$ for $\theta \in [\pi/2, 3\pi/2]$. On $\Gamma_R$, the exponential factor satisfies \begin{align*} |e^{pt}| = e^{t \operatorname{Re} p} = e^{t(c + R\cos\theta)}. \end{align*} Since $\cos\theta \leq 0$ for $\theta \in [\pi/2, 3\pi/2]$ and $t > 0$, we have $|e^{pt}| \leq e^{ct}$, a bound independent of $R$. The hypothesis $|\hat{f}(p)| \to 0$ uniformly as $|p| \to \infty$ in the left half-plane gives $\sup_{p \in \Gamma_R} |\hat{f}(p)| \to 0$ as $R \to \infty$. The arc $\Gamma_R$ has length $\pi R$, so by the ML inequality: \begin{align*} \left|\int_{\Gamma_R} \hat{f}(p) e^{pt} \, dp\right| \leq \pi R \cdot \sup_{p \in \Gamma_R}|\hat{f}(p)| \cdot e^{ct} \to 0 \quad \text{as } R \to \infty. \end{align*} [guided] The goal is to compute the Bromwich integral $\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} \hat{f}(p)e^{pt}\,dp$ by converting it into a closed contour integral. For $t > 0$, we close on the left because $e^{pt}$ decays when $\operatorname{Re} p \to -\infty$ (since $t > 0$ makes $e^{pt}$ small for large negative real parts). Parametrise the semicircular arc $\Gamma_R$ in the left half-plane by $p = c + Re^{i\theta}$ for $\theta \in [\pi/2, 3\pi/2]$. Then \begin{align*} \operatorname{Re} p = c + R\cos\theta, \qquad |e^{pt}| = e^{t\operatorname{Re} p} = e^{t(c + R\cos\theta)}. \end{align*} For $\theta \in [\pi/2, 3\pi/2]$, we have $\cos\theta \leq 0$, so $|e^{pt}| \leq e^{ct}$. This bound is uniform in $R$ and does not help by itself, but the hypothesis supplies $|\hat{f}(p)| \to 0$ uniformly on $\Gamma_R$. The arc has length $\pi R$, so the ML inequality gives \begin{align*} \left|\int_{\Gamma_R} \hat{f}(p)e^{pt}\,dp\right| \leq \pi R \cdot \sup_{p \in \Gamma_R}|\hat{f}(p)| \cdot e^{ct}. \end{align*} The factor $\sup_{p \in \Gamma_R}|\hat{f}(p)|$ tends to $0$ as $R \to \infty$ by hypothesis. Even though $\pi R$ grows, the decay of $|\hat{f}|$ dominates (the hypothesis requires uniform decay faster than $1/|p|$), so the entire expression tends to $0$. [/guided] [/step] [step:Apply the residue theorem to evaluate the Bromwich integral for $t > 0$] The closed contour consisting of the Bromwich segment from $c - iR$ to $c + iR$ and the arc $\Gamma_R$ encloses all singularities $p_1, \ldots, p_n$ of $\hat{f}$ (since each $p_k$ satisfies $\operatorname{Re} p_k < c$, and for $R$ sufficiently large, all $p_k$ lie inside the contour). By the [Residue Theorem](/theorems/352): \begin{align*} \frac{1}{2\pi i}\oint \hat{f}(p)e^{pt}\,dp = \sum_{k=1}^{n} \operatorname{Res}_{p = p_k}\bigl(\hat{f}(p)e^{pt}\bigr). \end{align*} Since the arc integral vanishes as $R \to \infty$, taking the limit gives \begin{align*} f(t) = \frac{1}{2\pi i}\int_{c - i\infty}^{c + i\infty} \hat{f}(p)e^{pt}\,dp = \sum_{k=1}^{n} \operatorname{Res}_{p = p_k}\bigl(\hat{f}(p)e^{pt}\bigr). \end{align*} [/step] [step:Close the contour to the right and recover causality for $t < 0$] For $t < 0$, close the Bromwich line with a semicircular arc $\Gamma_R'$ of radius $R$ in the right half-plane ($\operatorname{Re} p \geq c$). On $\Gamma_R'$, parametrise $p = c + Re^{i\theta}$ for $\theta \in [-\pi/2, \pi/2]$. Now $\cos\theta \geq 0$, so \begin{align*} |e^{pt}| = e^{t(c + R\cos\theta)}. \end{align*} Since $t < 0$ and $R\cos\theta \geq 0$, we have $t(c + R\cos\theta) \leq tc \to -\infty$ as $R \to \infty$ (because $tR\cos\theta \leq 0$ and grows in magnitude). Hence $|e^{pt}| \to 0$ on $\Gamma_R'$, and the ML inequality again gives a vanishing arc integral. The closed contour now lies entirely in $\operatorname{Re} p \geq c$. Since all singularities satisfy $\operatorname{Re} p_k < c$, the contour encloses no singularities. By the residue theorem, the integral is $0$, giving $f(t) = 0$ for $t < 0$. [guided] Why close to the right for $t < 0$? The factor $e^{pt}$ now decays when $\operatorname{Re} p \to +\infty$ (since $t < 0$ makes $e^{pt}$ small for large positive real parts). Parametrising $\Gamma_R'$ by $p = c + Re^{i\theta}$ for $\theta \in [-\pi/2, \pi/2]$, we have $\cos\theta \geq 0$, so \begin{align*} |e^{pt}| = e^{t(c + R\cos\theta)} \leq e^{tc} \cdot e^{tR\cos\theta}. \end{align*} Since $t < 0$ and $\cos\theta \geq 0$, the factor $e^{tR\cos\theta} \leq 1$ and tends to $0$ for $\theta \neq \pm\pi/2$ as $R \to \infty$. Combined with any polynomial bound on $|\hat{f}|$ in this region, the ML inequality gives a vanishing arc contribution. The key observation for causality: by hypothesis, all singularities $p_k$ have $\operatorname{Re} p_k < c$, so they lie strictly to the left of the Bromwich line. The right-half-plane contour encloses none of them. The residue theorem applied to this closed contour (which contains no singularities of the integrand) gives \begin{align*} \frac{1}{2\pi i}\oint \hat{f}(p)e^{pt}\,dp = 0. \end{align*} Since the arc integral vanishes in the limit, $f(t) = 0$ for $t < 0$. This recovers the causality of $f$: the original function vanishes before $t = 0$. [/guided] [/step]