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Let $f(t)$ be a function (for example of time $t$).

Is there a general expression of the laplace transform of $\sqrt{f(t)}$ ?

Same question for the inverse Laplace transform : Let $f(s)$ be the Laplace transform of $f(t)$, is there a generic expression of the inverse Laplace transform of $\sqrt{f(s)}$ ?

edit: One special case is known: the Laplace transform of $f(t)=t$. It is often tabulated, like in this table from Lamar University, Texas.

edit2: I have no idea how to prove such a general transform exists. Perhaps it does not and in this case I would be very grateful for a link or an explaination.

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You can start by trying special cases like $\sqrt{t}$, $\sqrt{e^t}$. But, for example,$\sqrt{\sin t}$ would be undefined. – Pedro Tamaroff Feb 5 '12 at 19:24
Why should there be? – Fabian Feb 5 '12 at 19:38

1 Answer 1

I am afraid there is no general relation for the laplace transform of a function.

The table you presented is quite rich in relations for well behaved families, such as $f_{a,2n}(t) = t^{2n} e^{at}$ since $\sqrt{f_{a,2n}} = f_{a/2,n}$ we get the following relation: $$ F_{\sqrt{f_{a,2n}}}(s) = F_{f_{a/2,n}}(s) = \frac{n!}{(s-a/2)^{n+1}}.$$ Moreover, if you consider functions that are constants, then you get a different relaion $F_{\sqrt{c}}(s) = \frac{\sqrt{c}}{s}$. Another different relation might be obtained if you consider a positive step functions $\phi(t) = \sum_{j=1}^k c_k 1_{[a_k,b_k)}(t)$ ($[a_k,b_k)$ disjoint ) then $\sqrt{\phi}(t)$ is again a simple function, so you can find a relation in this case too. $$F_{\sqrt{\phi}}(s) = \sum_k \sqrt{c_k}\frac{ e^{-s b_k } - e^{-sa_k}}{s}$$ All efforts seem unable to treat a general case, but you can infer from this different examples that such relation will always need some ad hoc consideration (that is, specific for each case)

Moreover, if your function is not positive, then you might have a hard time to define the square root of $f$. That is a bit discouraging

To conclude this lines, let me try to offer you a general relation that might help your efforts:

Let $f(t)>0$ for every $t$

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

$$ F_{\sqrt{f}}(s)^2 \leq \frac{1}{s}\int_0^\infty s e^{-st} f(t)\, dt = \frac{1}{s}F_f(s)$$

where the inequality follows from Jensen's inequality ($\phi(\int f d\mu) \leq \int \phi(f) d\mu$ as long as $\phi$ is convex and $\mu$ is a positive measure with total mass $1$ and note that $\int_0^\infty s e^{-st} = 1$)

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