# Proof of inverse Laplace transform

Why is

$$f(t) = \frac{1}{2πj}\int_{\sigma-j\infty}^{\sigma+j\infty} F(s) e^{st} \, ds,$$

provided that

$$F(s) = \int_{0}^{\infty} f(t) e^{-st} \, dt \ ?$$

I tried to find out myself, or searched online and found a term Bromwich integral, but I want to know how this expression is derived. (And I couldn't find any :()

Thank you.

• You missed the factor $\frac{1}{2\pi i}$ in front of the Bromwich integral. Sep 22, 2018 at 17:50
• @SangchulLee Oh, thank you, edited. Sep 22, 2018 at 23:22
• For the History of this, which is substantial, I suggest reading Operational Methods in Applied Mathematics by H. S. Carslaw. It's an inexpensive Dover publication now. Here's a link at Amazon: amazon.com/Operational-methods-applied-mathematics-advanced/dp/… Oct 7, 2018 at 17:56
• @DisintegratingByParts Thank you! I will add that to my to-do list after CSAT. Oct 7, 2018 at 22:14

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.)

• Thank you for your elaborate answer! Sep 23, 2018 at 6:19
• @sangchullee An integrable function $f(x)$ need not approach $0$ as $x\to \infty$. So, do we need an additional condition on $f$ here? Mar 24, 2019 at 19:41
• @MarkViola, That's another place the integrability of $f'(x)e^{-\sigma x}$ enters. Indeed, notice that $$f(b)e^{-\sigma b} - f(a)e^{-\sigma a} = \int_{a}^{b} \left( f(t) e^{-\sigma t} \right)' \, \mathrm{d}t = \int_{a}^{b} (f'(t) - \sigma f(t)) e^{-\sigma t} \, \mathrm{d}t$$ and that $t \mapsto (f'(t) - \sigma f(t)) e^{-\sigma t}$ is integrable. So, as $a, b \to \infty$, this converges to zero, and by the Cauchy criterion, $f(x)e^{-\sigma x}$ converges as $x \to \infty$. The limiting value is automatically determined as $0$ by the integrability of $f(x)e^{-\sigma x}$. Mar 24, 2019 at 19:45
• To simplify a bit, let's take $f(x)=e^{\sigma x}h(x)$. Then, by assumption $h$ is integrable. So, by the Cauchy Criterion, $\lim_{a,b\to\infty}\int_a^b h'(x)\,dx=\lim_{a,b\to\infty}(h(b)-h(a))=0$. While it is evident that $\lim_{x\to \infty}h(x)$ exists since $\lim_{b\to \infty}\int_a^b h'(x)\,dx=\lim_{b\to \infty}(h(b)-h(a))$ exists, how does one conclude that $\lim_{x\to \infty}h(x)=0$. Mar 24, 2019 at 21:13
• @MarkViola, The trick is rather simple: if $\lim_{x\to\infty}h(x)$ is other than zero, then $h$ cannot be integrable on $(0, \infty)$. Mar 24, 2019 at 21:26

Here I present another version of the inversion formula for the Laplace Transform based entirely on Fourier transform. This version extends the version described by Sangchul Lee.

Thought this posting

• $$m$$ denotes the Lebesgue measure on the real line.
• For any function $$f\in L^{loc}_1((0,\infty)$$, its Laplace transform is defined as $$\bar{f}(s)=\int^\infty_0e^{-st}f(t)\,dt$$ If the integral above converges absolutely for some $$s>0$$, then $$\bar{f}$$ can be extended as an analytic function of the half place $$H_{s}=\{z\in\mathbb{C}:\mathfrak{R}(z)>s\}$$ that is continuous along the vertical line $$\mathfrak{R}(z)=s$$.
• For any $$g\in L_1(\mathbb{R})$$, its Fourier transform is defined as $$\widehat{g}(\xi)=\int_\mathbb{R} e^{-2\pi I\xi x}g(x)\,dx$$ For any complex Borel measure (or real valued measure of total finite variation) $$\mu$$ on $$\mathbb{R}$$, its Fourier transform (or its characteristic function) is defined as $$\widehat{\mu}(\xi)=\int_\mathbb{R}e^{ix\xi}\mu(dx)$$ Notice that if $$\mu\ll m$$ and $$\mu=g\cdot m$$, then $$\widehat{g}(\xi)=\widehat{\mu}(-2\pi \xi)$$.

Inversion formulas for the Fourier transform are well known and have been discussed here at MSE. I will use the following version of the Fourier inversion formula.

Theorem ($$L_1$$ summability) Suppose $$f\in L_1(m)$$ and that $$f$$ is of bounded variation in a neighborhood of some point $$x\in\mathbb{R}$$. Then \begin{align} \frac{f(x-)+f(x+)}{2}=\frac{1}{2\pi}\lim_{T\rightarrow\infty}\int^T_{-T}e^{-itx} \widehat{f}(-t/2\pi)\,dt\tag{4}\label{four} \end{align} where $$f(x-)=\lim_{y\nearrow x}f(y)$$ and $$f(x+)=\lim_{y\searrow x}f(y)$$.

I provide a proof of this result at the end of my posting.

For example, if $$f\in L_1(m)$$ is piecewise continuously differentiable, then $$f$$ is local bounded variation and thus, \eqref{four} holds.

Derivation of inversion formula for Laplace Transform: Extend $$f$$ to $$\mathbb{R}$$ by setting $$f(t)=0$$ for $$t\leq 0$$. Suppose $$g_c(t)=e^{-ct}f(t)\in L_1((0,\infty),m)$$ for some $$c>0$$. It follows that $$\widehat{g_c}(\xi)=\int^\infty_{0}e^{-(c+2\pi\xi i)t}f(t)\,dt=\overline{f}(c+2\pi \xi i)$$ Then, by the summability theorem above, $$\frac{g_c(t-)+g_c(t_+)}{2}=\frac1{2\pi}\lim_{T\rightarrow\infty}\int^T_{-T}e^{-it\xi}\widehat{g_c}(-\xi/2\pi)\,d\xi=\frac1{2\pi}\lim_{T\rightarrow\infty}\int^T_{-T}e^{it\xi}\overline{f}(c+i\xi)\,d\xi$$ at any point $$t$$ around which $$g_c$$ (equivalently $$f$$) is of bounded variation. Hence, for such point $$t$$, $$\frac{f(t-)+f(t-)}{2}=\frac1{2\pi}\lim_{T\rightarrow\infty}\int^T_{-T}e^{t(c+i\xi)}\overline{f}(c+i\xi)\,d\xi= \frac{1}{2\pi i}\lim_{T\rightarrow\infty}\int^{c+iT}_{c-iT}e^{tz}\overline{f}(z)\,dz$$

Proof of $$L_1$$ summability theorem: By Fubini's theorem we have that $$g(x):=\int^T_{-T}\widehat{f}(-t/2\pi) e^{-ixt}\,dt=\int_{\mathbb{R}}\int^T_{-T}f(y)e^{i(y-x)t}\,dt\,dy= \int_{\mathbb{R}}f(y)\frac{\sin(T(y-x))}{y-x}dy$$ Since $$t\mapsto\frac{\sin t}{t}$$ is even, it follows that $$g(x)=\int_{\mathbb{R}}f(y+x)\frac{\sin(Ty)}{y}dy=\int_{\mathbb{R}}f(x-y)\frac{\sin(Ty)}{y}dy$$

Suppose $$f$$ is of bounded variation in the interval $$I_\delta=(x-\delta,x+\delta)$$. With our loss of generality, we may assume that $$f$$ is monotone nondecreasing on $$I_\delta$$. Splitting the domain of integration gives \begin{align} g(x)&=\frac1\pi\int_{\mathbb{R}}\frac{f(y+x)+f(x-y)}{2}\frac{\sin Ty}{y}\,dy=\frac2\pi\int^\infty_0\frac{f(y+x)+f(x-y)}{2}\frac{\sin Ty}{y}\,dy\\ &=\frac2\pi\int^\delta_0\left(\frac{f(y+x)+f(x-y)}{2}-\frac{f(x-)+f(x+)}{2}\right)\frac{\sin Ty}{t}\,dy\\ &\quad + \frac2\pi\int^\infty_\delta\frac{f(y+x)+f(x-y)}{2}\frac{\sin Ty}{y}\,dy\\ &\quad + \frac{f(x-)+f(x+)}{2}\frac{2}{\pi}\int^\delta_0\frac{\sin Ty}{y}\,dy \end{align}

The third integral in the right converges to $$\frac{f(x-)+f(x+)}{2}\frac{2}{\pi}\int^\infty_0\frac{\sin u}{u}\,du=\frac{f(x-)+f(x+)}{2}$$.

The second integral in the right converges to $$0$$ by The Riemann-Lebesgue lemma, for $$y\mapsto\frac{f(y+x)-f(x-y)}{y}\mathbb{1}_{(\delta,\infty)}(y)\in L_1(m)$$.

The first integral on the right requires a little extra effort. Let $$A=\sup_{y>0}\Big|\int^y_0\sin\frac{\sin y}{y}\,dy\Big|$$. Define \begin{align} h(y;x):=\frac{f(y+x)+f(x-y)}{2}-\frac{f(x-)+f(x+)}{2}= \frac{f(y+x)-f(x+)}{2} +\frac{f(x-y)-f(x-)}{2} \end{align}

Given $$\varepsilon>0$$, there is $$0<\eta<\delta$$ such that $$f(y)-f(x_+)<\frac{\varepsilon}{2A}$$ for $$x, and $$f(x-)-f(y)<\frac{\varepsilon}{2A}$$ for $$x-\delta.
By the second mean value theorem for integrals there are points $$\xi_+, \xi_-\in(0,\eta)$$ such that \begin{align} \int^\eta_0h(y;x)\frac{\sin Ty}{y}\,dy&=\frac{f(\xi_+ x)-f(x-)}{2}\int^\eta_{\xi_+}\frac{\sin Ty}{y}\,dy+\frac{f(x-\xi_-)-f(x-)}{2}\int^\eta_{\xi_-}\frac{\sin Ty}{y}\,dy \end{align} It follows that $$\Big|\int^\eta_0h(y;x)\frac{\sin Ty}{y}\,dy\Big|<\varepsilon$$ On the other hand, $$y\mapsto h(y;x)\mathbb{1}_{(\eta,\delta)}(y)\in L_1(m)$$ and so, $$\int^\delta_\eta h(y;x)\frac{\sin Ty}{y}\,dy$$ converges to $$0$$ by the Riemann-Lebesgue lemma.

Putting things together we have that $$\limsup_{T\rightarrow\infty}\left|\int^T_{-T}\widehat{f}(-t/2\pi) e^{-ixt}\,dy - \frac{f(x-)+f(x-)}{2}\right|\leq\varepsilon$$ As $$\varepsilon>0$$ is arbitrary, the conclusion of the Theorem follows.