Solve Inverse Laplace Transform Using Input Integral Theorem

Question

Problem

Using the input integral principle below

$$ \mathscr{L} \left[ \int_{0}^{t} f(u)du \right] (s) = \frac{1}{s} \mathscr{L} \left[ f(t) \right] (s), \ s > c $$

Find $ \mathscr{L}^{-1} \left[ \frac{1}{s(s^2+1)} \right](t) \ $ without using partial fractions.

Attempt

Letting $ \ f(t) = \mathscr{L}^{-1} \left[ \frac{1}{s(s^2+1)} \right](t) $,

$$ \mathscr{L} \left[ \int_{0}^{t} f(u)du \right] (s) = \frac{1}{s} \mathscr{L} \left[ \mathscr{L}^{-1} \left[ \frac{1}{s(s^2+1)} \right](t) \right] (s), \ s > c $$

$$ \mathscr{L} \left[ \int_{0}^{t} f(u)du \right] (s) = \frac{1}{s} \left[ \frac{1}{s(s^2+1)} \right], \ s > c $$

$$ \mathscr{L} \left[ \int_{0}^{t} f(u)du \right] (s) = \frac{1}{s^2(s^2+1)} , \ s > c $$

Notes

Perhaps I've approached this problem incorrectly, but I'm confused as how to proceed with it. All I'm looking for is a hint or correct first step in solving this problem, with a little bit of explanation as to what the correct method toward solving this problem entails.

Also, I searched up what the input integral principle is and I'm not finding anything on it. Did my professor invent this name or is it an alias for something else?

That being said, any help is appreciated. Thanks!

Answer 1

Compare these $$ \mathscr{L} \left[ \int_{0}^{t} f(u)du \right] (s) = \frac{1}{s} \mathscr{L} \left[ f(t) \right] (s) $$ $$ \mathscr{L} \left[ \int_{0}^{t} \color{red}{\sin u} \ du \right] (s) = \frac{1}{s} \color{red}{\frac{1}{(s^2+1)}} $$

Answer 2

Hint: The implied way is $$ \mathscr{L} \left[ \int_{0}^{t} f(u)du \right] (s) = \frac{1}{s} \mathscr{L} \left[ f(t) \right] (s) = \frac{1}{s}\frac{1}{s^2+1} \implies \mathscr{L} \left[ f(t) \right] (s) = \frac{1}{s^2+1}.$$

Answer 3

Note that your $$f(t)= \sin t $$ so $$ \mathscr{L}^{-1} \left[ \frac{1}{s(s^2+1)} \right](t)=\int _0^t \sin u du = 1- \cos t$$