The statement is wrong in implying anything particular about ``time domain'' and ``frequency domain''. The equality F(t)=F(s) is utterly trivial, since the name of the variable doesn't matter, what matters is that the interval of its values is the domain of the function.The function doesn't ``know'' that t is time and s is frequency, or whatever.

The correct and interesting question to ask is, for which functions f(x), F(t)=Integrate(exp(-t*x)*f(x),x=0,oo)=f(t). We integrate over x, so the result is a function of t. Since t does take values in the domain of f, it does make sense to ask for what functions does f(t) equal its transform at the same point.

This is a standard exercise about the fixed point(s) of the operator Integrate(exp(-t*x) ( ), x=0,oo) that maps functions to functions.

The only ``benefit'' is a deeper understanding of the action of such operators. But the statements have to make sense first.

Dear Stam Nicolis

You have overlooked F(t), wrongly as f(t).

You mentioned :" Since t does take values in the domain of f, it does make sense to ask for what functions does f(t) equal its transform at the same point."

Notice: I did not claim f(t) equals its transform at the same point. Whereas, I claimed the numerical value of the function F(.), is equivalent in Laplace-variable domain and in time domain; F(t)=F(s). Please notice that F(t) is not f(t). Please discriminate between them as one is of capital letter, and the other is of small letter.

Once more, that F(t)=F(s) is true by definition, since t and s take values in the same domain. Just changing the name of a variable doesn’t change anything.

F(t) = f(t) is the more interesting equality, because it isn't true by definition, but holds for certain functions.

Dear Stam Nicolis

There is no guarantee that F(t)=f(t). However, my conclusion that F(t)=F(s), could be interesting, if we think deeply. This is not only a change of the variable-name of the functions arguments. It says, the numerical value of the Laplace-transform of a function; F(s), at a specific value of the variable; s, is equivalent to its numerical value when we substitute directly the corresponding value of t for s, in F(s).

It seems the functions are equivalent when t and s approach infinity, it means simply; F(s)=F(t), when (s,t)→(∞,∞). They could be also approximatively the same for the all real line.

Moreover, the Laplace-variable; s, and the time-variable; t, could be both an algebraic value, although Laplace-variable could be further a complex or imaginary number, while time-variable; t, is always a positive and real number.

Nor does the name of the variable in the integral matter:

F(t)=Integrate(exp(-x*t)*f(x),x=0,oo)=Integrate(exp(-y*t)*f(y),y=0,oo)

Both define the Laplace transform of f. As does

F(s)=Integrate(exp(-z*s)*f(z),z=0,oo)

The Laplace transform of a function doesn't depend on what its argument is called. That's why

F(t)=Integrate(exp(-x*t)*f(x),x=0,oo)=Integrate(exp(-y*t)*f(y),y=0,oo)=F(s)=Integrate(exp(-x*s)*f(x),x=0,oo)

In the last equality the only change is the change in the name of the independent variable.

In these equalities F(s)=F(t) ALWAYS, by definition-they just represent two different names for the same point on the real line.

It's just a change in convention and nothing more. In all cases the domain of f is the real line and the domain of F is, also, the real line.

The reason the Laplace transform is useful is because it defines a particular mapping between functions. The properties of that mapping are of interest, not the names of variables. What they're called doesn't matter. By some convention the mapping is between f(t) and F(s); it could equally well be defined between f(x) and F(y) and nothing would change.

Dear Stam Nicolis

Although the both variable-names; s, and t, give the same numerical value of the integral, but there is a precious symmetry property hidden in the statement.

Consider the integral:

F(s)=L{f(t)}=integrate(exp(-cos(s*t))*f(t)*dt, t=0...∞).

If we replace s for t, then this integral is not numerically equivalent to the integral:

F(t)=L{f(t)}=integrate(exp(-cos(s^2))*f(s)*ds, s=0...∞).

where, cos, denotes cosine.

Hence, the conclusion is not true for any integral by assuming the change of integration variable. But the precious property is due to the mirror-like symmetry of exp(-st), and the integrand. This conclusion could be fruitful in some aspects.

Of course not, because, once more, the variable that's integrated over, doesn't appear outside the integral!

The second expression is meaningless.

The function; F(t), is an expression in time-domain, whist, the Function; F(s), is an expression in s-domain. If those are numerically equivalent, then it would have some significant implications, to relate time and frequency domains. This is NOT simply a change of variables for function arguments. Meanwhile, s could be frequency while t is the explicit time-variable and theses are widely far apart. Therefore, to simply interconnect them through the statement which I have already accomplished, could be of considerable importance, particularly when dealing with transfer-functions in control systems.

Update:

I mention your viewpoint a few lines above, directly quoted as:

"The correct and interesting question to ask is, for which functions f(x), F(t)=Integrate(exp(-t*x)*f(x),x=0,oo)=f(t). We integrate over x, so the result is a function of t. Since t does take values in the domain of f, it does make sense to ask for what functions does f(t) equal its transform at the same point."

You are right, and I will investigate thoroughly:

for what functions does f(t) equal its transform at the same point (?)

Although I already have a glimpse into the answer, based on time-frequency symmetry (duality).

Please see my ResearchGate project:

https://www.researchgate.net/project/Analytic-solution-to-the-Blasius-equation

and the PDF update with the title:

Analytic solution to the Blasius equation,

where you may see my idea to relate inverse-Laplace transform to time-frequency symmetry, to ultimately find a simple expression for f(t), knowing already the F(s).