Is f(x) of exponential order for the Laplace Transform?

[EngWiPy]

Hello,

By definition, the forward Laplace transform of a function $$f(x)$$ is:

$$\mathcal{L}\left\{f(x)\right\}=\int_0^{\infty}\text{e}^{-sx}f(x)\,dx$$.

Can we say the same for the function $$f\left(\frac{1}{x}\right)$$, i.e.:

$$\mathcal{L}\left\{f\left(\frac{1}{x}\right)\right\}=\int_0^{\infty}\text{e}^{-\frac{s}{x}}f\left(\frac{1}{x}\right)\,dx$$.??

Thanks in advance

 

[LCKurtz]

No, at least not as you have written it. If $g(x) = f(1/x)$ then

$$\mathcal{L}(g) = \int_0^\infty e^{-sx}g(x)\, dx = \int_0^\infty e^{-sx}f(1/x)\, dx$$

assuming g(x) is of exponential order.

 

[EngWiPy]

No, at least not as you have written it. If $g(x) = f(1/x)$ then

$$\mathcal{L}(g) = \int_0^\infty e^{-sx}g(x)\, dx = \int_0^\infty e^{-sx}f(1/x)\, dx$$

assuming g(x) is of exponential order.

Ok, I see. But what do you mean by "exponential order"?

 

[LCKurtz]

Ok, I see. But what do you mean by "exponential order"?

A function f is of exponential order if, informally, it grows no faster than an exponential. The definition is there exists a > 0 such that:

$$\lim_{t \rightarrow \infty} |f(t)e^{-at}| = 0$$

This is the usual condition given to ensure the Laplace transform of f exists.

 

[EngWiPy]

A function f is of exponential order if, informally, it grows no faster than an exponential. The definition is there exists a > 0 such that:

$$\lim_{t \rightarrow \infty} |f(t)e^{-at}| = 0$$

This is the usual condition given to ensure the Laplace transform of f exists.

Ok, thanks.

Regards