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Suppose we know the Laplace transform of a function $f(t)$:

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

Can we connect this to the Laplace transform of $h(t) = f(t^2)$? Is there a useful relation here?

Generalization: This is related to https://math.stackexchange.com/a/803458/10063, and http://eprints.cs.univie.ac.at/4962/1/2006_-_A_spectral_analysis_of_function_composition.pdf, which asks a more general question about Fourier transforms. I wonder if a similar method can be applied to Laplace transform. In general, we want to know the Laplace transform of $h(t) = f(g(t))$, if we know the Laplace transforms of $f(t)$ and $g(t)$.

a06e
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1 Answers1

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You can change variables $t^2=x$:

$$ H(s) = \int_0^\infty f(t^2) e^{-st}\; dt = \int_0^\infty \frac{f(x)}{2\sqrt{x}} e^{-s\sqrt{x}}\; dx $$ Now we can express $e^{-s \sqrt{x}}/\sqrt{x}$ as a Laplace transform, so: $$ \eqalign{ H(s) &= \int_0^\infty \int_0^\infty \frac{f(x)}{2 \sqrt{\pi t}} e^{-s^2/(4 t)} e^{-xt}\; dt\; dx \cr &= \int_0^\infty\frac{F(t)}{2 \sqrt{\pi t}} e^{-s^2/(4t)}\; dt}$$

Robert Israel
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  • It does not answer to the question but with similar argument, it seems that we have $H(s) = \int_{0}^{\infty} f(t^3) e^{-st} dt = \frac{\sqrt{s}}{3\pi} \int_{0}^{\infty} \frac{F(t)}{\sqrt{t}} K_{\frac{1}{3}} \Bigg( \frac{2}{3} \frac{s^{3/2}} {\sqrt{3t}} \Bigg) dt$ where $K$ is the modified Bessel function. – edrezen Nov 17 '23 at 22:23
  • And for the initial question: $H(s) = \int_{0}^{\infty} f(t^2) e^{-st} dt = \frac{{s}}{2^{3/2}\pi} \int_{0}^{\infty} \frac{F(t)}{{t}} K_{1/2} \Big( \frac{s^{2}} {{4t}} \Big) dt $ – edrezen Nov 18 '23 at 12:45
  • Maybe there exists some kind of generalization for $n>3$ involving the modified Bessel function but it is not obvious at first glance. – edrezen Nov 18 '23 at 12:52