# What is the Inverse Laplace Transform of e^(-sx^2/2)?

• Haku
In summary: I see. So what is the easier way to calculate the inverse?The easier way to calculate the inverse is to use the time-shift property of Laplace transforms and shift the function to the right by the appropriate amount.
Haku
Homework Statement
Find the inverse Laplace transform of 1/s^2 * e^(-sx^2/2)
Relevant Equations
Laplace convolution theorem equation.
My attempt at finding this was via convolution theorem, where we take F(s) = 1/s^2 and G(s) = e^(-sx^2/2). Then to use convolution we need to find the inverses of those transforms. From a table of Laplace transforms we know that f(t) = t. But I am sort of struggling with e^(-sx^2/2). My 'guess' is that the inverse Laplace transform of e^(-sx^2/2) is δ(t - (x^2)/2). This is from the fact that the inverse Laplace transform of e^sc is δ(t+c). But then integrating this in the convolution integral proves difficult.
Am I on the right track or should I use another method of finding the inverse Laplace transform of 1/s^2 * e^(-sx^2/2)?
Note: Here I have used * to denote multiplication and not convolution.

It seems a bit puzzling to me that you say the presence of the Dirac delta function makes evaluating the integral difficult. I find it makes evaluating integrals simple and straightforward. Can you elaborate more about why you find it difficult?

Also, are you wedded to using convolution? A straightforward application of the established properties of Laplace transforms will get you the answer to the problem.

Haku
vela said:
It seems a bit puzzling to me that you say the presence of the Dirac delta function makes evaluating the integral difficult. I find it makes evaluating integrals simple and straightforward. Can you elaborate more about why you find it difficult?

Also, are you wedded to using convolution? A straightforward application of the established properties of Laplace transforms will get you the answer to the problem.
I am happy to use any method, preferably the easiest too.
The reason I said it made it difficult is because you would have to evaluate the Dirac delta function at ((t-T)-x^2/2) would you? (Where T is tau).

Don't let the argument of the delta function confuse you. You have
$$\delta\left[(t-\tau) - \frac{x^2}{2}\right] = \delta\left[\left(t - \frac{x^2}{2}\right)-\tau\right].$$ You can treat the stuff in the parentheses as a constant since you're integrating with respect to ##\tau##.

Also, you should be a little careful here. The inverse Laplace transform of ##F(s)## is ##f(t) = t u(t)## where ##u(t)## is the unit step function.

Haku
vela said:
Don't let the argument of the delta function confuse you. You have
$$\delta\left[(t-\tau) - \frac{x^2}{2}\right] = \delta\left[\left(t - \frac{x^2}{2}\right)-\tau\right].$$ You can treat the stuff in the parentheses as a constant since you're integrating with respect to ##\tau##.

Also, you should be a little careful here. The inverse Laplace transform of ##F(s)## is ##f(t) = t u(t)## where ##u(t)## is the unit step function.
Ah okay, I will try to integrate that now.
As for F(s), why would it be tu(t)? the Laplace transform of t is 1/s^2. So why would the inverse of that change the original function?
Also what was your easier way to calculate the inverse?
Thank you!

Haku said:
As for F(s), why would it be tu(t)? the Laplace transform of t is 1/s^2. So why would the inverse of that change the original function?
If you recall, when you calculate ##F(s)##, you calculate
$$F(s) = \int_0^\infty t\,e^{-st}dt.$$ All information about ##f(t)## when ##t<0## doesn't contribute to ##F(s)##. Hence, when you say the inverse Laplace transform of ##F(s) = 1/s^2## is ##f(t) = t##, it's really only true for ##t>0##. Quite often, this detail doesn't matter as it's understood that ##t>0## (for example, when solving a differential equation). In problems like this, where the factor ##e^{-as}## represents a shift in time by ##a##, it becomes important to remember that ##f(t) = 0## for ##t<0##. When you shift the function to the right by ##a##, the result is ##0## for ##t<a##. (Note this is the simpler way: use the time-shift property of Laplace transforms.)

Another way to see this is to try calculating the Laplace transforms of
$$f(t) = \begin{cases} t-a, & t \ge a \\ 0, & t<a \end{cases}$$ and
$$g(t) = t-a$$ where ##t>0##. You'll see ##f## yields the Laplace transform ##e^{-as}/s^2## whereas ##g## doesn't.

Haku
vela said:
If you recall, when you calculate ##F(s)##, you calculate
$$F(s) = \int_0^\infty t\,e^{-st}dt.$$ All information about ##f(t)## when ##t<0## doesn't contribute to ##F(s)##. Hence, when you say the inverse Laplace transform of ##F(s) = 1/s^2## is ##f(t) = t##, it's really only true for ##t>0##. Quite often, this detail doesn't matter as it's understood that ##t>0## (for example, when solving a differential equation). In problems like this, where the factor ##e^{-as}## represents a shift in time by ##a##, it becomes important to remember that ##f(t) = 0## for ##t<0##. When you shift the function to the right by ##a##, the result is ##0## for ##t<a##. (Note this is the simpler way: use the time-shift property of Laplace transforms.)

Another way to see this is to try calculating the Laplace transforms of
$$f(t) = \begin{cases} t-a, & t \ge a \\ 0, & t<a \end{cases}$$ and
$$g(t) = t-a$$ where ##t>0##. You'll see ##f## yields the Laplace transform ##e^{-as}/s^2## whereas ##g## doesn't.
Ahhhh yes, it is because the lower limit of the integral is 0 right?

So in this case you get; $$F(s) = \int_0^\infty t\,e^{-st}dt.$$ = 1/s^2
And we have the s-shifting variable e^-as, where as = sx^2/2 right?
But in my definition of the s-shifting prop we used e^at as the shifting 'variable', in this case we have t = 1 but also an x term so how does that all fit together?

Also, do we consider the case of t < 0 because once it is shifted, then it is no long t < 0, it becomes t < a, which could be > 0 right? Since the actual t < 0 would not matter since t denotes time right?

Thanks!

There are two shifting properties: one in the time domain and one in the s-domain. You're mixing them up.

Even with the shift, your solution needs to be defined for ##0 \le t \lt a##. That's why the unit step function is included.

Haku
vela said:
There are two shifting properties: one in the time domain and one in the s-domain. You're mixing them up.

Even with the shift, your solution needs to be defined for ##0 \le t \lt a##. That's why the unit step function is included.
Ah yes I see,

I think I understand now, so by the second shifting property we have that the inverse Laplace transform of e^(-sc)F(s) = f(t-c)H(t-c), and in this example c = x^2/2 and F(s) = 1/s^2.

So we get the inverse Laplace transform of e^(-sx^2/2)•1/s^2 = f(t-c)H(t-c), where c = x^2/2 and H is the unit step function.
That means the original function u(x, t) = the inverse Laplace transform of e^(-sx^2/2)•1/s^2 = f(t-x^2/2)H(t-x^2/2) = (t-x^2/2)u(t-x^2/2) right?
Thanks!

Haku said:
That means the original function u(x, t) = the inverse Laplace transform of e^(-sx^2/2)•1/s^2 = f(t-x^2/2)H(t-x^2/2) = (t-x^2/2)u(t-x^2/2) right?
It's not correct to write "e^(-sx^2/2)•1/s^2 = f(t-x^2/2)H(t-x^2/2)" but you got the idea.

Haku
vela said:
It's not correct to write "e^(-sx^2/2)•1/s^2 = f(t-x^2/2)H(t-x^2/2)" but you got the idea.
Ah I see.
Thank you! Helped a lot!

## What is an inverse Laplace transform?

The inverse Laplace transform is a mathematical operation that allows us to find the original function from its Laplace transform. It is the reverse process of the Laplace transform and is used to solve differential equations and analyze dynamic systems.

## How is the inverse Laplace transform calculated?

The inverse Laplace transform is calculated using the Bromwich integral, which is a complex contour integral. It involves integrating the Laplace transform function along a contour in the complex plane and then taking the limit as the contour approaches infinity. This process can be done analytically or numerically using software or tables.

## What are the applications of the inverse Laplace transform?

The inverse Laplace transform has many applications in physics, engineering, and mathematics. It is commonly used to solve differential equations in fields such as electrical circuits, control systems, and signal processing. It is also used in probability and statistics to analyze stochastic processes.

## What is the relationship between the Laplace transform and the inverse Laplace transform?

The Laplace transform and the inverse Laplace transform are inverse operations of each other. The Laplace transform converts a function from the time domain to the frequency domain, while the inverse Laplace transform converts it back from the frequency domain to the time domain. They are used together to solve differential equations and analyze dynamic systems.

## Are there any limitations to using the inverse Laplace transform?

Yes, there are some limitations to using the inverse Laplace transform. It is not always possible to find the inverse Laplace transform of a function, especially if the function is complex or has poles on the imaginary axis. In these cases, numerical methods or approximations may be used. Additionally, the inverse Laplace transform may not exist for functions that grow too quickly or are not well-behaved.

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