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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.

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  • $\begingroup$ You missed the factor $\frac{1}{2\pi i}$ in front of the Bromwich integral. $\endgroup$ – Sangchul Lee Sep 22 '18 at 17:50
  • $\begingroup$ @SangchulLee Oh, thank you, edited. $\endgroup$ – KYHSGeekCode Sep 22 '18 at 23:22
  • $\begingroup$ 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/… $\endgroup$ – DisintegratingByParts Oct 7 '18 at 17:56
  • $\begingroup$ @DisintegratingByParts Thank you! I will add that to my to-do list after CSAT. $\endgroup$ – KYHSGeekCode Oct 7 '18 at 22:14
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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.)

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    $\begingroup$ Thank you for your elaborate answer! $\endgroup$ – KYHSGeekCode Sep 23 '18 at 6:19
  • $\begingroup$ @sangchullee An integrable function $f(x)$ need not approach $0$ as $x\to \infty$. So, do we need an additional condition on $f$ here? $\endgroup$ – Mark Viola Mar 24 at 19:41
  • $\begingroup$ @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} $. $\endgroup$ – Sangchul Lee Mar 24 at 19:45
  • $\begingroup$ 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 $\lim_{x\to \infty}h(x)=0$. $\endgroup$ – Mark Viola Mar 24 at 21:13
  • $\begingroup$ @MarkViola, The trick is rather simple: if $\lim_{x\to\infty}h(x) $ is other than zero, then $h$ cannot be integrable on $(0, \infty)$. $\endgroup$ – Sangchul Lee Mar 24 at 21:26

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