Real Integration Theorem
This theorem establishes the relationship between the Laplace transform of a function and that of its integral.
It states that
$$
\mathscr{L}\left[ \int_0^t f(t) \,\text{d}t \right]
= \frac{1}{s} F(s)
$$
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.
\begin{align*}
&\, \int_0^\infty \int_0^t f(t)
e^{-st}(-s) \left( -\frac{1}{s} \right)
\,\text{d}t \,\text{d}t
\\
&\left(
u = \int_0^t f(t) \, \text{d}t,
\quad
\text{d}v = -s e^{-st} \, \text{d}t
\right)
\\
=&\, – \frac{1}{s}
\left.
\int_0^t f(t) e^{-st} \,\text{d}t \,
\right|_{0}^\infty
+ \frac{1}{s}
\int_0^\infty f(t) e^{-st} \, \text{d}t
\\
=&\, f(\infty) e^{-s \infty} \frac{1}{-s} + f(0) + \frac{1}{s} F(s)
= 0 + \frac{1}{s} F(s)
\end{align*}
(Original image here.)
Best Answer
Correct, but use different variables:
$$\mathscr{L}\left[ \int_0^t f(\tau) \, d\tau\right] = \int_0^{\infty} e^{-st} \int_0^t f(\tau) \, d\tau \, dt. $$
We have:
$$ \int_0^{\infty} e^{-st} \int_0^t f(\tau) \, d\tau \, dt = \int_0^{\infty} s\,e^{-st} \int_0^t \frac{1}{s} f(\tau) \, d\tau \, dt. $$
Applying the famous technique, integration by parts:
$$dv = s\, e^{-st} dt \quad \Rightarrow \quad v = -e^{-st}, $$ $$u = \int_0^t \frac{1}{s} f(\tau) \, d\tau \quad \Rightarrow \quad du=\frac{1}{s} f(t), $$
so:
$$\int_0^{\infty} s\,e^{-st} \int_0^t \frac{1}{s} f(\tau) \, d\tau \, dt. = - \frac{1}{s} e^{-st} \left. \int_0^t f(\tau)\, d\tau \, \right|_{0}^\infty + \frac{1}{s} \int_0^\infty f(t)\, e^{-st} \, dt$$
$$= \frac{1}{s} \int_0^\infty f(t)\, e^{-st} \, dt$$
$$ = \frac{1}{s}\, F(s).$$
Now, we add a second integral:
$$\mathscr{L}\left[ \int_0^t \int_0^{\sigma} f(\tau) \, d\tau \, d\sigma\right] = \int_0^{\infty} e^{-st} \left( \int_0^t \int_0^{\sigma} f(\tau) \, d\tau \, d\sigma \right) dt. $$
Again, we use integration by parts:
$$ \int_0^{\infty} e^{-st} \left( \int_0^t \int_0^{\sigma} f(\tau) \, d\tau \, d\sigma \right) dt = \int_0^{\infty} s\, e^{-st} \left( \int_0^t \int_0^{\sigma} \frac{1}{s} f(\tau) \, d\tau \, d\sigma \right) dt, $$
with:
$$dv = s\, e^{-st} dt \quad \Rightarrow \quad v = -e^{-st}, $$ $$u = \int_0^t \int_0^{\sigma} \frac{1}{s} f(\tau) \, d\tau \, d\sigma \quad \Rightarrow \quad du= \int_0^t \frac{1}{s} f(\tau)\, d\tau. $$
Well, this is long:
$$\int_0^{\infty} s\, e^{-st} \left( \int_0^t \int_0^{\sigma} \frac{1}{s} f(\tau) \, d\tau \, d\sigma \right) dt = - \frac{1}{s} e^{-st} \left. \int_0^t \int_0^{\sigma} f(\tau) \, d\tau \, d\sigma \, \right|_{0}^\infty + \int_0^{\infty} e^{-st} \int_0^t \frac{1}{s} f(\tau)\, d\tau, $$
$$ = \frac{1}{s} \int_0^{\infty} e^{-st} \int_0^t f(\tau)\, d\tau, $$ $$ = \frac{1}{s}\cdot \frac{1}{s} F(s). $$