# Solving Laplace Transform of (1-3cost)/t^2

• sara_87
In summary: The question is asking for [1 - 3 cos(t)]/t^2;i was thinking exactly the same. i mean what integral limits do we put to actually evaluate the laplace transform? if the integral limits are s and infinity then log term in s(ln(s)) - s - (3/2)s(ln(s^2+1)) + (3s-arctan(s)) would diverge.

## Homework Statement

Find the laplace transform of the following:
(1-3cost)/t^2

## The Attempt at a Solution

I know many methods to find the laplace transform like the first and second shift theorems, the differentiation one, convolution, but for this particular question i don't know where to start. Any hints would be very much appreciated. thank you.

Start with the Laplace transform of 1 - 3 cos(t). Then integrate both sides of the integral w.r.t. s twice.

oh right so i integrate: (1/s) - (3s)/(s^2+1) twice wrt s.
once gives:
ln(s)-3/2(ln(s^2+1))

so i integrate this again using integration by parts to give:
s(ln(s)) - s - (3/2)s(ln(s^2+1)) + (3s-arctan(s))

is that right?

It looks right, but you need toi ceck the problem again. Is it really the Laplace transform of [1 - 3 cos(t)]/t^2 that is wanted? If so, how is this Laplace transorm to beinterpreted, as the ordinary Laplace integral does not converge (due to the singularity in the integrand at t = 0).

This gives rise to the asymptotic s Log(s), behavior, which you cannot get if you compute a Laplace integral that converges, because if you let s go to infinity, the laplace transorm should go to zero.

If you had [1 - cos(t)]/t^2, then the result would be slightly different and you would find that in the large s expansion, the s Log(s) terms cancel (as well as the log(s) and s terms, only terms that approach zero for s to infinity remain).

Last edited:
The question is asking for [1 - 3 cos(t)]/t^2;
i was thinking exactly the same. i mean what integral limits do we put to actually evaluate the laplace transform? if the integral limits are s and infinity then log term in s(ln(s)) - s - (3/2)s(ln(s^2+1)) + (3s-arctan(s)) would diverge.

?

sara_87 said:
The question is asking for [1 - 3 cos(t)]/t^2;
i was thinking exactly the same. i mean what integral limits do we put to actually evaluate the laplace transform? if the integral limits are s and infinity then log term in s(ln(s)) - s - (3/2)s(ln(s^2+1)) + (3s-arctan(s)) would diverge.

?

Yes, clearly the integral of exp(-st) [1-3cos(t)]/t^2 dt from zero to infinity does not converge. However, you can imagine doing something like the following. We have a map L(f) that maps functions for which we can define the Laplace integral to their Laplace transform (as defined by that Laplace integral). Then, we can try to extend L(f) to a larger domain of functions for which the Laplace integral is not defined using some of the properties that L(f) satisfies when the domain is restricted to the Laplace integrable case.

which properties? sorry, i don't understand what you mean. so does that mean we have to choose the limits?

## 1. What is the purpose of solving Laplace Transform of (1-3cost)/t^2?

The purpose of solving the Laplace Transform of (1-3cost)/t^2 is to transform a function from the time domain to the frequency domain. This allows for easier analysis and solving of differential equations.

## 2. What is the Laplace Transform of (1-3cost)/t^2?

The Laplace Transform of (1-3cost)/t^2 is equal to 3/s^2 - 1/s.

## 3. How do you solve the Laplace Transform of (1-3cost)/t^2?

To solve the Laplace Transform of (1-3cost)/t^2, you can use the definition of the Laplace Transform or look up the transform in a table. You may also use properties of the Laplace Transform, such as linearity and time shifting.

## 4. What are the steps involved in solving the Laplace Transform of (1-3cost)/t^2?

The steps involved in solving the Laplace Transform of (1-3cost)/t^2 include breaking the function into simpler components, applying any necessary properties of the Laplace Transform, and then using integration techniques to find the transform.

## 5. What are some applications of the Laplace Transform of (1-3cost)/t^2?

The Laplace Transform of (1-3cost)/t^2 has various applications in engineering, physics, and mathematics. It is commonly used in solving differential equations, analyzing electrical circuits, and studying the behavior of systems over time.