MHB Existence of Laplace transform

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The discussion centers on proving that if a function f is piecewise continuous on [0,∞) and of exponential order c, then the integral of e^(-st)f(t) from 0 to ∞ is analytic in the right half-plane where Re(s) > c. A key point raised is the incorrect assumption that the absolute value of e^(-st) equals e^(-st) for complex s, which was acknowledged as a significant mistake in the proof. The correction emphasizes the need to properly handle the complex nature of s in the analysis. The overall focus is on establishing the conditions under which the Laplace transform is valid and analytic. This discussion highlights the importance of careful mathematical reasoning in proofs related to Laplace transforms.
alyafey22
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Prove the following

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

$$\int^\infty_0 e^{-st} f(t)\, dt $$​

is analytic in the right half-plane for $$\mathrm{Re}(s)>c$$
 
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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 $\text{Re}(s) > c$.
 
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ZaidAlyafey said:
Prove the following

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

$$\int^\infty_0 e^{-st} f(t)\, dt $$​

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

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

$\chi$ $\sigma$
 
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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.
 
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