Calculate Laplace transform of funcion f(x)=1

Question

i have to calculate transforma of Laplace of the function $f(x)=1$.

By definition we have $$ L(1)(s)= \displaystyle\int_0^{+\infty} e^{-sx} dx= \lim_{A \to +\infty} \displaystyle\int_0^A e^{-s x} dx. $$ I found that $L(1)(s)= \dfrac{1}{s}$ where $Re(s)>0$.

My question is how we find this result? please

Answer 1

The laplace transform is defined as

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

Here $f(t) = 1$

$L\{1\} = \int_0^\infty e^{-st}\,dt$

$L\{1\} = \lim_{A\to\infty}\bigg(\frac {-e^{-st}}s\bigg|_0^A\bigg)$

$L\{1\} = \lim_{A\to\infty}\frac{-e^{-At}}s+\frac {e^0}s$

$L\{1\} = \frac 1s$ $\,\,\,\,$ provided $s\gt 0$

as $\lim_{A\to\infty}e^{-A} = 0$

Answer 2

it is a fundamental property of the number $e$ that: $$ \frac{d}{dx} e^x = e^x $$

from this, and using the elementary formula for differentiating a function of a function we have: $$ \frac{d}{dx} e^{f(x)} = f'(x) e^{f(x)} $$

so if $f(x) = -sx$, we have $f'(x) = -s$, giving: $$ \frac{d}{dx} e^{-sx} = -s e^{-sx} $$

which may be written as

$$ e^{-sx} = \frac1{s} \frac{d}{dx} e^{-sx} $$

this gives:

$$ \int_0^A e^{-sx} \, dx = \int_0^A -\frac1{s} \frac{d}{dx} (e^{-sx}) dx = -\frac1{s} \int_0^A \frac{d}{dx}(e^{-sx}) dx = \bigg[-\frac1{s}e^{-sx} \bigg]_0^A = \frac1{s} ( 1 - e^{-As}) $$

if $A \gt 0$ and $s \gt 0$ then $ 0 \lt e^{-s} \lt 1$ so $e^{-As } = (e^{-s})^A \to 0$ as $A \to +\infty$