Vtyjkyuk

Real Integration Theorem

This theorem establishes the relationship between the Laplace transform of a function and that of its integral.
It states that
$$
mathscrLleft[ int_0^t f(t) ,textdt right]
= frac1s F(s)
$$
The proof of this theorem is carried out by integrating the definition of the Laplace transform by parts.
This proof is similar to that of the real differentiation theorem and is left as an exercise.
The Laplace transform of the $n$th intgegral of a function is the transform of the function divides by $s^n$.

(Original image here.)

Hello. I would like to check whether my solving is right or not in proving the rule stated in the picture above.
Also I would like to know how is it solved for a second integral $(n=2)$? just to be convinced with the general rule of $n$th integral.
Thank you.

beginalign*
&, int_0^infty int_0^t f(t)
e^-st(-s) left( -frac1s right)
,textdt ,textdt
\
&left(
u = int_0^t f(t) , textdt,
quad
textdv = -s e^-st , textdt
right)
\
=&, - frac1s
left.
int_0^t f(t) e^-st ,textdt ,
right|_0^infty
+ frac1s
int_0^infty f(t) e^-st , textdt
\
=&, f(infty) e^-s infty frac1-s + f(0) + frac1s F(s)
= 0 + frac1s F(s)
endalign*

(Original image here.)

Correct, but use different variables:

$$mathscrLleft[ int_0^t f(tau) , dtauright]
= int_0^infty e^-st int_0^t f(tau) , dtau , dt. $$

We have:

$$ int_0^infty e^-st int_0^t f(tau) , dtau , dt = int_0^infty s,e^-st int_0^t frac1s f(tau) , dtau , dt. $$

Applying the famous technique, integration by parts:

$$dv = s, e^-st dt quad Rightarrow quad v = -e^-st, $$
$$u = int_0^t frac1s f(tau) , dtau quad Rightarrow quad du=frac1s f(t), $$

so:

$$int_0^infty s,e^-st int_0^t frac1s f(tau) , dtau , dt. = - frac1s e^-st left. int_0^t f(tau), dtau , right|_0^infty + frac1s int_0^infty f(t), e^-st , dt$$

$$= frac1s int_0^infty f(t), e^-st , dt$$

$$ = frac1s, F(s).$$

Now, we add a second integral:

$$mathscrLleft[ int_0^t int_0^sigma f(tau) , dtau , dsigmaright]
= int_0^infty e^-st left( int_0^t int_0^sigma f(tau) , dtau , dsigma right) dt. $$

Again, we use integration by parts:

$$ int_0^infty e^-st left( int_0^t int_0^sigma f(tau) , dtau , dsigma right) dt = int_0^infty s, e^-st left( int_0^t int_0^sigma frac1s f(tau) , dtau , dsigma right) dt, $$

with:

$$dv = s, e^-st dt quad Rightarrow quad v = -e^-st, $$
$$u = int_0^t int_0^sigma frac1s f(tau) , dtau , dsigma quad Rightarrow quad du= int_0^t frac1s f(tau), dtau. $$

Well, this is long:

$$int_0^infty s, e^-st left( int_0^t int_0^sigma frac1s f(tau) , dtau , dsigma right) dt = - frac1s e^-st left. int_0^t int_0^sigma f(tau) , dtau , dsigma , right|_0^infty + int_0^infty e^-st int_0^t frac1s f(tau), dtau, $$

$$ = frac1s int_0^infty e^-st int_0^t f(tau), dtau, $$
$$ = frac1scdot frac1s F(s). $$