Calculating the Inverse Laplace Transform for a Given Function

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Discussion Overview

The discussion revolves around finding the inverse Laplace transform of the function $F(s)=\frac{1}{s(s^2+1)}$. Participants explore methods for calculating the inverse transform, including the use of partial fraction decomposition and reference to tables of Laplace transforms.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • Participants express a desire to find $f(t)$ given the Laplace transform $F(s)$ and discuss the formula for the inverse Laplace transform.
  • Some participants suggest using partial fraction decomposition to simplify the expression $\frac{1}{s(s^2+1)}$ into $\frac{1}{s}-\frac{s}{s^2+1}$.
  • There is a reference to a table of Laplace transforms to identify the transforms of $\frac{1}{s}$ and $\frac{s}{s^2+1}$.
  • One participant notes that the inverse Laplace transform of $\frac{1}{s}$ is $u(t)$, and questions how this relates to the overall solution.
  • Another participant states that the inverse Laplace transform of $\frac{s}{s^2+1}$ is $\cos(t) \cdot u(t)$ and questions the implications of these results.
  • Participants derive that $f(t) = 1 - \cos(t)$ from the inverse transforms and discuss the relevance of the Heaviside function in this context.
  • There is a clarification about the relationship between $f(t)$ and $F(s)$, emphasizing that the inverse Laplace transform is defined such that $\mathscr L[f(t)](s) = F(s)$.
  • One participant mentions the formula in post #1 as a solution found as Mellin's inverse formula or the Bromwich integral.

Areas of Agreement / Disagreement

Participants generally agree on the method of using partial fraction decomposition and the results from the Laplace transform table, but there is some confusion regarding the implications of these results and the role of the Heaviside function. The discussion remains exploratory without a definitive conclusion.

Contextual Notes

Participants express uncertainty about how to proceed with the integral and the implications of the inverse Laplace transform results. There are unresolved questions about the necessity of the Heaviside function in the final expression for $f(t)$.

evinda
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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)
 
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evinda said:
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)
 
I like Serena said:
Hey evinda! (Smile)

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

It holds that $\frac{1}{s(s^2+1)}=\frac{1}{s}-\frac{s}{s^2+1}$.
I like Serena said:
And then look up the resulting fractions in a table. (Thinking)

What do you mean? (Thinking)
 
evinda said:
It holds that $\frac{1}{s(s^2+1)}=\frac{1}{s}-\frac{s}{s^2+1}$.

Good! (Happy)

evinda said:
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)
 
I like Serena said:
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)
 
evinda said:
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)
 
I like Serena said:
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)
 
evinda said:
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)
 
I like Serena said:
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)
 
  • #10
evinda said:
Ah I see... (Nod)

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

Yep.

evinda said:
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)
 
  • #11
I like Serena said:
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)
 

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