MHB Calculating the Inverse Laplace Transform for a Given Function

[evinda]

Hello! (Wave)

I want to find $f(t)$ if its Laplace transform is $F(s)=\frac{1}{s(s^2+1)}$.

We use the following formula, right?

$$f(t)=\frac{1}{2 \pi i} \lim_{T \to +\infty} \int_{a-iT}^{a+iT} e^{st} F(s) ds$$

But how can we calculate the integral $\int_{a-iT}^{a+iT} e^{st} \frac{1}{s(s^2+1)} ds$ ? (Thinking)

 

[I like Serena]

Hello! (Wave)

I want to find $f(t)$ if its Laplace transform is $F(s)=\frac{1}{s(s^2+1)}$.

We use the following formula, right?

$$f(t)=\frac{1}{2 \pi i} \lim_{T \to +\infty} \int_{a-iT}^{a+iT} e^{st} F(s) ds$$

But how can we calculate the integral $\int_{a-iT}^{a+iT} e^{st} \frac{1}{s(s^2+1)} ds$ ? (Thinking)

Hey evinda! (Smile)

The usual way is to do a partial fraction decomposition first.
And then look up the resulting fractions in a table. (Thinking)

 

[evinda]

Hey evinda! (Smile)

The usual way is to do a partial fraction decomposition first.

It holds that $\frac{1}{s(s^2+1)}=\frac{1}{s}-\frac{s}{s^2+1}$.

And then look up the resulting fractions in a table. (Thinking)

What do you mean? (Thinking)

 

[I like Serena]

It holds that $\frac{1}{s(s^2+1)}=\frac{1}{s}-\frac{s}{s^2+1}$.

Good! (Happy)

What do you mean? (Thinking)

Take a look at this List of Laplace Transforms.
Can we find $\frac 1s$ and $\frac s{s^2+1}$ in it? (Wondering)

 

[evinda]

Good! (Happy)
Take a look at this List of Laplace Transforms.
Can we find $\frac 1s$ and $\frac s{s^2+1}$ in it? (Wondering)

We have the integral $\int_{a-iT}^{a+iT} \left( \frac{e^{st}}{s}-\frac{s e^{st}}{s^2+1}\right) ds$, don't we?

So how does it help to find the Laplace tranform of $\frac 1s$ and $\frac s{s^2+1}$? I am confused right now... (Worried)

 

[I like Serena]

We have the integral $\int_{a-iT}^{a+iT} \left( \frac{e^{st}}{s}-\frac{s e^{st}}{s^2+1}\right) ds$, don't we?

So how does it help to find the Laplace tranform of $\frac 1s$ and $\frac s{s^2+1}$? I am confused right now... (Worried)

Isn't according to that table for instance $\mathscr L^{-1}\left[\frac 1s\right](t) = u(t)$? (Wondering)

We can verify by evaluating:
$$\mathscr L[u(t)](s) = \int_0^\infty u(t)e^{-st}\,dt
= \int_0^\infty e^{-st}\,dt = -\frac 1s e^{-st}\Big|_0^\infty = \frac 1s
$$
can't we? (Wondering)

 

[evinda]

Isn't according to that table for instance $\mathscr L^{-1}\left[\frac 1s\right](t) = u(t)$? (Wondering)

We can verify by evaluating:
$$\mathscr L[u(t)](s) = \int_0^\infty u(t)e^{-st}\,dt
= \int_0^\infty e^{-st}\,dt = -\frac 1s e^{-st}\Big|_0^\infty = \frac 1s
$$
can't we? (Wondering)

Ok... And $\mathscr L^{-1}\left[\frac{s}{s^2+1}\right](t)=\cos{t} \cdot u(t)$, where $u$ is the Heaviside function. Right?But what do we have from that? (Thinking)

 

[I like Serena]

Ok... And $\mathscr L^{-1}\left[\frac{s}{s^2+1}\right](t)=\cos{t} \cdot u(t)$, where $u$ is the Heaviside function. Right?But what do we have from that? (Thinking)

Yep.
It means that:
$$f(t)=\mathscr L^{-1}[F(s)](t) = \mathscr L^{-1}\left[\frac 1s - \frac s{s^2+1}\right](t) = u(t) - \cos t\cdot u(t) = 1-\cos t
$$
We can leave out the Heaviside step function, since the domain of $t$ would be $\mathbb R_{\ge 0}$. (Thinking)

 

[evinda]

Yep.
It means that:
$$f(t)=\mathscr L^{-1}[F(s)](t) = \mathscr L^{-1}\left[\frac 1s - \frac s{s^2+1}\right](t) = u(t) - \cos t\cdot u(t) = 1-\cos t
$$
We can leave out the Heaviside step function, since the domain of $t$ would be $\mathbb R_{\ge 0}$. (Thinking)

Ah I see... (Nod)

The fact that $f(t)=\mathscr L^{-1}[F(s)](t) $ is known, right? If we would want to prove it, would we take the formula of post #1? (Thinking)

 

[I like Serena]

Ah I see... (Nod)

The fact that $f(t)=\mathscr L^{-1}[F(s)](t) $ is known, right?

Yep.

If we would want to prove it, would we take the formula of post #1? (Thinking)

As I understand it, it's really the other way around.
The inverse Laplace transform of $F(s)$ is $f(t)$ such that $\mathscr L[f(t)](s)=F(s)$ is satisfied.
Apparently the formula in post #1 is a solution found as Mellin's inverse formula, the Bromwich integral, or the Fourier–Mellin integral. (Thinking)

 

[evinda]

Yep.
As I understand it, it's really the other way around.
The inverse Laplace transform of $F(s)$ is $f(t)$ such that $\mathscr L[f(t)](s)=F(s)$ is satisfied.
Apparently the formula in post #1 is a solution found as Mellin's inverse formula, the Bromwich integral, or the Fourier–Mellin integral. (Thinking)

I see... Thanks a lot! (Smile)