Yesterday I had my first contact with the Laplace transform, in an Electric Circuits class. $\mathcal{L} \left\{ u \left( -t \right) \right\}$ showed up. Our teacher said it was equal to $\frac{1}{s}$, but couldn't tell exactly why.

I want to know if it's possible to explain why $\mathcal{L} \left\{ u \left( -t \right) \right\} = \frac{1}{s}$ by using known properties and transforms, like those found in Wikipedia's page about Laplace transform, as if I was trying to do it myself, considering I learned the very basics.

No, it's not for an assignment.

Thanks in advance.

Edit: $u \left( t \right)$ is the unit step function. $u \left( -t \right)$ was used because it was needed reversed.

  • $\begingroup$ $u(-t)$ or $u(t)$? Is $u$ the unit step function? $\endgroup$
    – JohnD
    Nov 29, 2014 at 17:06

4 Answers 4


I realize that this is an older question, but I'll answer it anyway in case other people are looking for the answer.

As an electrical engineer, a Laplace transform usually references the two-sided Laplace transform. Therefore the function being transformed does not have to be causal, meaning that the function can equal something other than $0$ for $t < 0$. (Here's a wiki reference to the two-sided Laplace transform: https://en.wikipedia.org/wiki/Two-sided_Laplace_transform )

Using this version of the Laplace transform,

$$\mathcal{L}\{ u(-t)\}(s) = \int_{-\infty}^\infty u(-t)e^{-st} \ dt = \int_{-\infty}^0 e^{-st} \ dt = -\frac{1}{s}$$

This is true only when $Re[s] < 0$, which is this transform's region of convergence.

I hope this helps!

Sources: I'm an electrical engineering student.

  • $\begingroup$ This is probably the reason of confusion. Nice first answer. Just one thing. You should have $\text{Re}\,s<0$, right? $\endgroup$
    – mickep
    Oct 5, 2015 at 5:03
  • $\begingroup$ Oops, thank you for pointing that out, I'll fix the mistake. $\endgroup$ Oct 5, 2015 at 5:05

I suspect you have a typo in what you posted since $u(-t)=0$ for $t>0$. Laplace transforms assume the underlying function is causal. Otherwise, computing the Laplace transform of the zero function is trivial.

Now, if you mean $u(t)$ (as I suspect) which is the unit step function, just compute its Laplace transform straightaway from the definition: $$ \mathscr{L}\{u(t)\}=\int_0^\infty e^{-st}u(t)\,dt=\int_0^\infty e^{-st}\cdot 1\,dt={e^{-st}\over -s}\Bigg|_{t=0}^{t=\infty}=-{1\over s}(0-1)={1\over s}, \quad s>0.$$

  • $\begingroup$ $u(-t)=0$ for $t\gt 0$ $\endgroup$
    – UserX
    Nov 29, 2014 at 17:11
  • $\begingroup$ I understand but I suspect his function has a typo since the Laplace transform of this is trivial and does not match his intended result. $\endgroup$
    – JohnD
    Nov 29, 2014 at 17:12
  • $\begingroup$ No, the laplace transform does not assume the underlying function is "causal". There's no typo. The laplace transform is not trivial in this case and actually matches his result (except possibly for the sign). $\endgroup$
    – skyking
    Oct 5, 2015 at 5:59

This looks wrong to me. Assuming by $u$ you mean the unit step function $u(t) = 1$ if $t \geq 0$ and $u(t) = 0$ otherwise, then by definition of the Laplace transform

$$\mathcal{L}\{ u(-t)\}(s) = \int_0^\infty u(-t)e^{-st} \ dt = \int_0^\infty 0 \ dt = 0$$

as for all $t \in [0,\infty)$, $u(-t) = 0$.

It is however true by the same sort of calculation that $\displaystyle \mathcal{L}\{ u(t)\}(s) = \frac{1}{s}$, as

$$\mathcal{L}\{ u(t)\}(s) = \int_0^\infty u(t)e^{-st} \ dt = \int_0^\infty e^{-st} \ dt = \frac{1}{s}$$

  • $\begingroup$ Relatedly, if you're looking for an introduction to Laplace transforms, I recommend these lectures from the ever amiable Arthur Mattuck of MIT: ocw.mit.edu/courses/mathematics/… (Lectures 19-23 of a course in ODEs. Discontinuous inputs like this one are dealt with in lecture 22.) $\endgroup$
    – Simon S
    Nov 29, 2014 at 17:15

Well he is wrong, the fact that he couldn't say why is a reason to mistrust his statement. You could come up with that result by using timescaling rule:

$\mathcal{L} f(a\cdot) = F(s/a)/a$

which with $a=-1$ would lead to the result that your teacher claims, but this rule assumes that $a>0$ so it can't be used. The reason the rule doesn't hold is that you in the process reverses the integration limits. If you put it into the formula you get:

$\mathcal{L} u(-\cdot) = \int_{-\infty}^{+\infty} u(-t)e^{-st}dt = \int_{-\infty}^0e^{-st}dt = [-s^{-1}e^{-st}]_{-\infty}^0 = -s^{-1}$ (if $\Re s < 0)$


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