Question

Inverse Laplace transform of "1" ?

Answer 1

Is it even possible because there is no "s" variable

Answer 2

Have you considered applying the definition?

Answer 3

I don't know how to calculate inverse laplace transforms without the table.

Answer 4

Just to be clear I mean \[\mathcal{L}^{-1}(1)\] and not\[\mathcal{L}(1)\]

Answer 5

We could use the definition of the Laplace transform if we let:
\( \mathcal{L}^{-1} (1) = f(t) \)
\( 1 = \mathcal{L} (f(t)) \)

Answer 6

so \(\mathcal{L}(\mathcal{L}^{-1}(1))=1\)... ? I'm not sure what to do =\ This just gives 1=1

Answer 7

The definition of Laplace transform of a function f(t):
\( \displaystyle \int_{0}^{\infty} e^{-st} f(t) \; \text{d}t = 1 \)

So, we'd be looking for a function f(t) that satisfies this equation...

Answer 8

oh I guess this could be transformed into the pdf of the gamma distribution?

Answer 9

this seems odd to me to use probability in my diff. equations class for which it is not even a prerequisite ._.

Answer 10

It is \(Dirac-\delta, \delta(t)\) or Unit Impulse at t = 0 - depending on the audience.

Answer 11

Oh interesting I googled this. Hm haven't encountered that yet. Maybe I made a mistake in my work.

Answer 12

oh yes I did actually! Hm hehe well thanks for the info :)!!!

Answer 13

I never fully understood this until just now I found this idea:
\( \displaystyle \int_{-\infty}^{\infty} \delta \left(t-c\right) \; \text{d}t = 1 \)
Although the domain of int would simply need to contain the value at x=c

Taking the laplace transform of \(\delta(t-c)\), we'd have:
\( \displaystyle \int_{0}^{\infty} e^{-st} \delta(t-c) \; \text{d}t \)
The area under the curve is essentially the value of the graph at x=c, since delta just eliminates the remaining graph and the only area taken is under the graph at that one point... \(\large = e^{-sc} \) We have c=0, so \(e^{-s0} = e^{0} = 1\)
Interesting. :D