Existence of Laplace transform

[alyafey22]

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 \(\displaystyle \mathrm{Re}(s)>c\)

 

[polygamma]

Spoiler

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 $\text{Re}(s) > c$.

 

[chisigma]

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 \(\displaystyle \mathrm{Re}(s)>c\)

Spoiler

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

$\chi$ $\sigma$

 

[polygamma]

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.

 

[CellCoree]

The Laplace transform is a powerful mathematical tool used in many areas of science and engineering. It allows us to transform a function from the time domain to the frequency domain, providing a new perspective on the behavior of the function. The existence of the Laplace transform has been proven through rigorous mathematical analysis.

In order to prove the given statement, we must first define what is meant by a function being of exponential order $c$. This means that there exists a constant $M$ such that $|f(t)| \leq Me^{ct}$ for all $t \geq 0$. This condition ensures that the integral in question is well-defined and converges.

Next, we must show that the integral $\int^\infty_0 e^{-st} f(t)\, dt$ is analytic in the right half-plane for $\mathrm{Re}(s)>c$. This means that the function is differentiable with respect to $s$ in this region and its derivative is continuous. This can be proven using the properties of the Laplace transform, such as linearity and differentiation in the time domain.

One key property that allows us to prove the analyticity of the Laplace transform is the Laplace inversion formula, which states that $f(t) = \frac{1}{2\pi i}\int_{\mathrm{Re}(s)=\sigma} e^{st}F(s)\, ds$, where $F(s)$ is the Laplace transform of $f(t)$ and $\sigma$ is any real number greater than the exponential order of $f(t)$. This formula allows us to express the original function $f(t)$ in terms of its Laplace transform $F(s)$, which is known to be analytic in the right half-plane for $\mathrm{Re}(s)>c$. Therefore, by the properties of analytic functions, $f(t)$ must also be analytic in this region.

In conclusion, the existence of the Laplace transform has been proven and the given statement is true. The Laplace transform is a valuable tool for solving differential equations and analyzing the behavior of systems in various fields of science and engineering. Its existence and properties make it a fundamental concept in mathematics and its applications continue to be explored and utilized in many different areas.