The Question

The inverse Laplace Transform formula can be read quite intuitively: $$ x(t) = \frac{1}{2\pi j}\int^{\sigma + j\infty}_{\sigma - j\infty} X(s)e^{st} ds $$ Given a particular damping factor $\sigma$, the signal in time domain is a weighted sum of eigenfunction $X(s)e^{st}$, across all frequency. The $\frac{1}{2\pi j}$ is just an overall scaling factor.

Now, how can I go from the inverse Laplace Transform formula to the Laplace Transform formula? (for example, the Bilateral Laplace Transform formula?) $$ X(s) = \int^{\infty}_{-\infty}x(t)e^{-st}dt $$

Supplementary Information

(Thank you Matt L. for answering this part)

I have tried to plug the Bilateral Laplace Transform formula into the Inverse Laplace Transform to see if I can obtain some insight, but after writing the initial step down, I don't know how should I proceed. $$ x(t) = \frac{1}{2\pi j}\int^{\sigma + j\infty}_{\sigma - j\infty} \left( \int^{\infty}_{-\infty}x(t)e^{-st}dt \right) e^{st} ds $$


1 Answer 1


I show you how an engineer would do it. First of all note that your final formula can be confusing because you overloaded the variable $t$. A better version is

$$\begin{align}x(t)&=\frac{1}{2\pi j}\int_{\sigma-j\infty}^{\sigma+j\infty}\left[\int_{-\infty}^{\infty}x(\tau)e^{-s\tau}d\tau\right]e^{st}ds\\\end{align}\tag{1}$$

If we set $s=\sigma+j\omega$ and integrate over $\omega$ you get

$$\begin{align}x(t)&=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}x(\tau)e^{-\sigma\tau}e^{-j\omega\tau} e^{\sigma t}e^{j\omega t}d\tau d\omega\\&=\frac{1}{2\pi}\int_{-\infty}^{\infty}x(\tau)e^{\sigma(t-\tau)}\int_{-\infty}^{\infty}e^{j\omega(t-\tau)}d\omega d\tau\\&=\frac{1}{2\pi}\int_{-\infty}^{\infty}x(\tau)e^{\sigma(t-\tau)}2\pi\delta(t-\tau) d\tau\\&=x(t)\end{align}\tag{2}$$

where I've used

$$\int_{-\infty}^{\infty}e^{j\omega t}d\omega=2\pi\delta(t)\tag{3}$$

with the Dirac delta impulse $\delta(t)$, and


which holds for any function $f(t)$ that is continuous at $t=t_0$.

  • $\begingroup$ Thanks for the help, your answer is nice and clear but I still have some problems. $\endgroup$
    – Eric
    Dec 30, 2017 at 22:10
  • $\begingroup$ First, my main problem is actually to get Bilateral Laplace Transform formula from the Inverse Laplace Transform formula and it is still not clear how I can do it. My last formula is just to show people how I attempted to attack the problem in case they are interested. Maybe I should have state my question more clearly. I will edit my question accordingly. $\endgroup$
    – Eric
    Dec 30, 2017 at 22:10
  • $\begingroup$ Secondly, in the 2nd line of (2) when you switched the integration order, how can I ensure it satisfy the Fubini's theorem? $\endgroup$
    – Eric
    Dec 30, 2017 at 22:10
  • $\begingroup$ @Eric: You can view the Laplace transform formula as a convolution of $e^{st}$ with $x(t)$ evaluated at $t=0$. If you consider $x(t)$ as an impulse response of a system, this convolution computes the eigenvalue $X(s)$. $\endgroup$
    – Matt L.
    Dec 31, 2017 at 8:34

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