Laplace Transform Definition

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Hello,

By definition, the forward Laplace transform of a function [tex]f(x)[/tex] is:

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

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

[tex]\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[/tex].??

Thanks in advance
 

LCKurtz

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

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

assuming g(x) is of exponential order.
 
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No, at least not as you have written it. If [itex] g(x) = f(1/x)[/itex] then

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

assuming g(x) is of exponential order.
Ok, I see. But what do you mean by "exponential order"?
 

LCKurtz

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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:

[tex] \lim_{t \rightarrow \infty} |f(t)e^{-at}| = 0[/tex]

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

[tex] \lim_{t \rightarrow \infty} |f(t)e^{-at}| = 0[/tex]

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

Regards
 

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