Proof of inverse Laplace transform

It is the Fourier inversion formula in disguise. In case you have never encountered this theorem before, let me prove the following version (which is obviously far from optimal).

Proposition. Let $F(s) = \int_{0}^{\infty} f(t)e^{-st} \, dt$ be the Laplace transform of $f : [0,\infty) \to \mathbb{R}$. Assume that the following technical conditions hold with some $g : [0,\infty) \to \mathbb{R}$ and $\sigma \in \mathbb{R}$:

• $f(t) = f(0) + \int_{0}^{t} g(u) \, du$. (In particular, $g$ is the 'derivative' of $f$.)
• Both $f(t)e^{-\sigma t}$ and $g(t)e^{-\sigma t}$ are Lebesgue-integrable on $[0, \infty)$.

Then for any $s > 0$, we have $$ \lim_{R\to\infty} \frac{1}{2\pi i} \int_{\sigma-iR}^{\sigma+iR} F(z)e^{s z} \, dz = f(s). $$

Proof. Define $S(x) = \frac{1}{2} + \frac{1}{\pi}\int_{0}^{x} \frac{\sin t}{t} \, dt$. Then $S(x)$ is bounded, and by Dirichlet integral, we have

$$ \lim_{R\to\infty} S(Rx) = H(x) := \begin{cases} 1, & x > 0 \\ \frac{1}{2}, & x = 0 \\ 0, & x < 0 \end{cases} $$

(Obviously $H$ denotes the Heaviside step function.) Now we have

\begin{align*} \frac{1}{2\pi i} \int_{\sigma-iR}^{\sigma+iR} F(z)e^{s z} \, dz &= \frac{1}{2\pi} \int_{-R}^{R} F(\sigma + i\xi)e^{s(\sigma+i\xi)} \, d\xi \\ &= \frac{1}{2\pi} \int_{-R}^{R} \left( \int_{0}^{\infty} f(t)e^{-(\sigma+i\xi)t} \, dt \right)e^{s(\sigma+i\xi)} \, d\xi. \end{align*}

By Fubini's theorem, we can interchange the order of integral to obtain

\begin{align*} \frac{1}{2\pi i} \int_{\sigma-iR}^{\sigma+iR} F(z)e^{s z} \, dz &= \int_{0}^{\infty} f(t)e^{-(t-s)\sigma} \left( \frac{1}{2\pi} \int_{-R}^{R} e^{(s-t)i\xi} \, d\xi \right) \, dt \\ &= \int_{0}^{\infty} f(t)e^{-(t-s)\sigma} \left( \frac{\sin R(t-s)}{\pi (t-s)} \right) \, dt \end{align*}

By the assumption, both $f(t)e^{-\sigma t}$ and $(f(t)e^{-\sigma t})' = (f'(t) - \sigma f(t))e^{-\sigma t}$ are Lebesgue-integrable. In particular, this tells that $f(t)e^{-\sigma t}$ converges to $0$ as $t\to\infty$. So by integration by parts,

\begin{align*} \frac{1}{2\pi i} \int_{\sigma-iR}^{\sigma+iR} F(z)e^{s z} \, dz &= - f(0)e^{s\sigma} S(-Rs) - \int_{0}^{\infty} (f(t)e^{-(t-s)\sigma})' S(R(t-s)) \, dt. \end{align*}

As $R \to \infty$, the right-hand side converges to

\begin{align*} \lim_{R\to\infty} \frac{1}{2\pi i} \int_{\sigma-iR}^{\sigma+iR} F(z)e^{s z} \, dz &= - \int_{0}^{\infty} (f(t)e^{-(t-s)\sigma})' H(t-s) \, dt \\ &= - \left[ f(t)e^{-(t-s)\sigma} \right]_{t=s}^{t=\infty} = f(s). \end{align*}

(Pushing the limit inside the integral is justified by the dominated convergence theorem.)