Is there a function that equals its Laplace transform?

[Bipolarity]

Google seems to provide not much information on this. In essence, I am asking about the eigenfunctions of the Laplace transform when λ=1? Anyone have any insights on this rather unusual problem?

BiP

 

[Simon Bridge]

You want: $$F(s) = \int_0^\infty f(t)e^{-st}dt = f(s)$$ ... for a function to be it's own Laplace transform.

Been discussed before:
https://www.physicsforums.com/archive/index.php/t-180250.html
I don't think there is any F(s)=f(s) ... but I'd be hard pressed to prove it.

 

[JJacquelin]

You want: $$F(s) = \int_0^\infty f(t)e^{-st}dt = f(s)$$ ... for a function to be it's own Laplace transform.
Been discussed before:
https://www.physicsforums.com/archive/index.php/t-180250.html
I don't think there is any F(s)=f(s) ... but I'd be hard pressed to prove it.

I agree in case of real function.
In case of complex function, there are many solutions. For example, one of them :

 

[Simon Bridge]

Yike - well there you have it :)

 

[blue_raver22]

olarBear, there is actually a function that equals its Laplace transform, known as the Dirac delta function (δ(t)). This function is defined as zero everywhere except at t=0, where it is infinitely tall and infinitely narrow, such that its area under the curve is equal to 1. Its Laplace transform is also equal to 1, and it is considered an eigenfunction of the Laplace transform with eigenvalue λ=1. However, it is important to note that the Dirac delta function is not a traditional function in the sense that it cannot be evaluated at any point except t=0. It is often used in mathematics and engineering as a convenient tool for representing impulsive phenomena, but it may not be applicable in all cases. I hope this helps answer your question.