# Need help compute the laplace transform

• slug
In summary, the conversation is about computing the Laplace transform of h(t) = -5t^2 and verifying the formula L[t^n] = n!/s^(n+1). The person had difficulty with the integration and got 10s instead of -10/s^3. They also mentioned using integration by parts and that it is a standard problem in ODE books.
slug
need help
compute the laplace transform of
h(t)= -5t^2
I took the integral((-5t^2)e^-(st)) i forgot integration by parts so i did it the hard way in my head and finally got
lim t -> oo 5[(-1/s)(t^2)(e^-st) - (2te^-st) - (2se^-st)] 0--oo
i got 10s but the book says -10/s^3
did i get the inetgration wrong or is it the second part that's wrong

i also have to verify that L[t^n]= n!/s^(n+1) basicaly the same problem but general
muchos gracias and stuff

$$\mathcal{L}\{-5t^2\}=-5\int_0^\infty e^{-st}t^2 dt$$

Let:

$$u=t^2$$

$$dv=e^{-st}dt$$

Really need to use parts for this. Also, this is a standard one that most ODE books will illustrate.

It seems like you may have made a mistake in your integration. The correct way to compute the Laplace transform of h(t) = -5t^2 would be to use the definition of the Laplace transform, which is the integral of e^(-st) multiplied by the function being transformed. In this case, it would be:

L[h(t)] = ∫e^(-st)(-5t^2)dt from 0 to infinity

To solve this integral, you can use integration by parts. Let u = -5t^2 and dv = e^(-st)dt. Then, du = -10t dt and v = (-1/s)e^(-st). Plugging this into the integration by parts formula, we get:

∫e^(-st)(-5t^2)dt = (-1/s)e^(-st)(-5t^2) from 0 to infinity - ∫(-1/s)e^(-st)(-10t)dt from 0 to infinity

= (-1/s)e^(-st)(-5t^2) + (10/s^2)e^(-st) from 0 to infinity

= (-5/s)e^(-st)t^2 + (10/s^2)e^(-st) from 0 to infinity

Now, taking the limit as t approaches infinity, we get:

lim t -> infinity (-5/s)e^(-st)t^2 + (10/s^2)e^(-st) = (10/s^2) - (5/s)(0) = 10/s^2

So the correct Laplace transform of h(t) = -5t^2 is 10/s^2, not 10s as you had calculated.

For the general case of L[t^n], we can use induction to prove that L[t^n] = n!/s^(n+1). First, we can easily verify that this formula holds for n = 0. L[t^0] = L[1] = 1/s = 0!/s^(0+1).

Now, let's assume that the formula holds for some arbitrary value of n. That is, L[t^n] = n!/s^(n+1). We want to show that this formula also holds for n+1. Using the definition of the Laplace transform, we have:

L[t^(n+1)] =

## 1. What is the Laplace Transform?

The Laplace Transform is a mathematical operation that converts a function of time into a function of complex frequency. It is commonly used in engineering and physics to solve differential equations and analyze systems in the frequency domain.

## 2. How is the Laplace Transform computed?

The Laplace Transform is computed by taking the integral of a function multiplied by the exponential function e^(-st), where s is a complex variable. This results in a new function that is a function of complex frequency.

## 3. What are the advantages of using the Laplace Transform?

The Laplace Transform has several advantages, including simplifying the process of solving differential equations, allowing for analysis of systems in the frequency domain, and providing a more intuitive representation of complex systems.

## 4. Can the Laplace Transform be used for any function?

In theory, the Laplace Transform can be used for any function that meets certain criteria, such as being absolutely integrable. However, in practice, it is typically used for functions that are piecewise continuous and have exponential order.

## 5. How is the Laplace Transform related to the Fourier Transform?

The Laplace Transform is closely related to the Fourier Transform, as it is a generalization of the Fourier Transform for complex functions. The Fourier Transform is a special case of the Laplace Transform when the complex variable s is set to 0.

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