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Laplace transform of $h(t)=f^{-1}(t)$ is:

$$L_s(h(t))=\int_0^\infty e^{-st}h(t)dt\mathop=^{h(t)\to t}\int_0^\infty t e^{-s f(t)} df(t)=\frac1s\int_0^\infty e^{-sf(t)}dt\mathop.$$

Could someone go through a step-by-step. Thanks

First part:

Using: $$h(t)=f^{-1}(t)$$ substituting: $u=f^{-1}(t)$ ->$f(u)=t$-> $df(u)=dt$

$$\int_0^\infty e^{-st}h(t)dt=\\\int_0^\infty e^{-st}f^{-1}(t)dt=\\\int_0^\infty e^{-sf(u)}f^{-1}(f(u))df(u)=\\\int_0^\infty e^{-sf(u)}u df(u)=\\\int_0^\infty te^{-sf(t)}df(t)=?$$

Mariusz Iwaniuk
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1 Answers1

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Assuming that $f$ is continuous and of local finite variation on $[0,\infty)$,

$$d_t(e^{-sf(t)})=-s e^{-sf(t)}\,df(t)$$

Integration by parts yields

$$\int_0^\infty t e^{-s f(t)} df(t)=-\frac1s\int^\infty_0td_t(e^{-sf(t)})=-\frac1s\Big(te^{-sf(t)}\Big|^\infty_0-\int^\infty_0e^{-sf(t)}dt\Big) $$

Under the assumption $\lim_{t\rightarrow\infty} te^{-sf(t)}=0$,

$$\int_0^\infty t e^{-s f(t)} df(t)=\frac1s\int^\infty_0 e^{-sf(t)}\,dt$$

Mittens
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