# A possibly illegal laplace transform appears to result in a derivative with an order $\lfloor x^2 \rfloor$? What does it mean? [closed]

I noticed that if I take the laplace transform of $x^{\lfloor s^2 \rfloor} y(x)$ that I get:

$(-1)^{\lfloor s^2 \rfloor} \frac {d^{\lfloor s^2 \rfloor}Y}{ds^{\lfloor s^2 \rfloor}}(s)$

This is based upon me noting that this is of the form $x^ny(x)$, where n is a constant integer function. Since "n" is independent of x, and floor returns integral values, I believe this is justified.

This prompts me to ask an obvious question:

What does it mean to take the derivative of order $\lfloor x^2 \rfloor$ with respect to x?

I imagine that such a derivative would just be each of the derivatives over piecewise intervals. I know that it's always integral valued, so there is no need to bring in fractional derivatives. Nothing in the notation seems to say that such a thing cannot occur, so it ultimately begs the question as to what it truly means and how it should be evaluated. I imagine that someone must've done this before now.

Of course, it might be illegal to place s within the function being transformed, in which case this question is pointless.

## closed as off-topic by Namaste, астон вілла олоф мэллбэрг, user91500, Gabriel Romon, user223391 Dec 10 '16 at 23:54

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• "Technically the floor function is always integral and in this sense we treat it as a constant." This is the fishy part. Obviously $x^{\lfloor x^2\rfloor}$ is not a constant. – Rahul Nov 26 '16 at 5:43
• I don't follow. Let's start with something simpler. The Laplace transform of a constant $n$ is $n/s$. The Laplace transform of $x$ is $1/s^2$. What should the Laplace transform of $\lfloor x\rfloor$ be, by your logic? (It is actually $1/((e^s-1)s)$.) – Rahul Nov 26 '16 at 6:08

## 1 Answer

The reason for the identity $\mathcal{L}\{t^nf(t)\} = F^{(n)}(s)$ is due to the fact that you can take the basic identity

$$F(s) = \int_0^\infty f(t) e^{-st} \,dt$$

and differentiate both sides $n$ times with respect to $s$. You can't do something similar to generate a $t^{\lfloor t^2 \rfloor}$ in front of your function, but what you can do is find the ranges over which $\lfloor t^2 \rfloor$ is constant, then break your function up into a piecewise sum of indicator functions times factors of the form $t^c f(t)$, then use linearity to evaluate the Laplace transform. I could only really see doing this for a particular function, not for a general function, where the Laplace transform would be very complicated. You'd probably be better off using power series methods or something of the like.

EDIT:

Answering the question at the end of your post, it is indeed illegal to place the ss inside the function being transformed. The function being transformed, f(t)f(t), is a function of one variable (in the frequency domain). Writing something like $\int_0^{\infty} e^{-st} t^{\lfloor s \rfloor} f(t) \,dt$ amounts to a new transform.

• Then answering the question at the end of your post, it is indeed illegal to place the $s$ inside the function being transformed. The function being transformed, $f(t)$, is a function of one variable (in the frequency domain). Writing something like $\int_0^\infty e^{-st} t^{\lfloor s \rfloor} f(t) \, dt$ amounts to a new transform. – CodeLabMaster Nov 26 '16 at 23:51