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Existence of Laplace transform

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Prove the following

Suppose that $f$ is piecewise continuous on \(\displaystyle [0,\infty) \) and of exponential order $c$ then

\(\displaystyle \int^\infty_0 e^{-st} f(t)\, dt \)​

is analytic in the right half-plane for \(\displaystyle \mathrm{Re}(s)>c\)
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
If $f(t)$ is of exponential order $c$, then there exists a real constant $c$ and positive constants $M$ and $T$ such that $|f(t)| \le M e^{c t}$ when $t > T$.

Then

$$|F(s)| = \Big| \int_{0}^{\infty} f(t) e^{-st} \ dt \Big| \le \int_{0}^{\infty} |f(t) e^{-st}| \ dt = \int_{0}^{T} |f(t) e^{-st} | \ dt + \int_{T}^{\infty} |f(t)e^{-st}| \ dt$$

$$ \le \int_{0}^{T} |f(t) e^{-st} | \ dt + M \int_{T}^{\infty} e^{ct} e^{-\text{Re}(s) t} \ dt $$

$$ = \int_{0}^{T} |f(t) e^{-st} | \ dt + M \int_{T}^{\infty} e^{[c-\text{Re}(s)]t} \ dt$$


The first integral converges for all values of $s$ since $f(t)$ is continuous.

And the second integral converges if $\text{Re} (s) > c $.

So $F(s)$ is absolutely convergent for $\text{Re}(s) >c$, and is thus complex differentiable (i.e., analytic) for $\text{Re}(s) > c$.
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
Prove the following

Suppose that $f$ is piecewise continuous on \(\displaystyle [0,\infty) \) and of exponential order $c$ then

\(\displaystyle \int^\infty_0 e^{-st} f(t)\, dt \)​

is analytic in the right half-plane for \(\displaystyle \mathrm{Re}(s)>c\)
A function f(t) is said to be of 'exponential order c' if for any M>0 exists a T>0 for which for all t>T is $\displaystyle |f(t)| \le M\ e^{c\ t}$. An f(t) of exponential order c admits Laplace Transform...

$\displaystyle \mathcal{L}\ \{f(t)\} = F(s) = \int_{0}^{\infty} f(t)\ e^{- s\ t}\ dt\ (1)$

... and the integral in (1) converges if $\text{Re}\ (s) > c$. Now applying the Inverse Laplace Transform formula to F(s) You have to obtain f(t) as follows...

$\displaystyle f(t) = \frac{1}{2\ \pi\ i}\ \int_{\gamma - i\ \infty}^{\gamma + i\ \infty} F(s)\ e^{s\ t}\ ds\ (2)$

... where $\gamma$ has to be $\ge c$ and on the right of all singularities of F(s) and that means that F(s) is analytic for all s for which is $\displaystyle \text{Re}\ (s) > c$...

Kind regards

$\chi$ $\sigma$
 
Last edited by a moderator:

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
I corrected a significant mistake in my proof.

I originally said that $|e^{-st}| = e^{-st}$.

That's obviously not true if $s$ is complex.