Question

Find the Laplace Transforms of
`\displaystyle{L\left \{\int_0^t t\cos 3t \, dt \right \}}`

Answer 1

To find the Laplace transform of
`\displaystyle{L\left \{\int_0^t t\cos 3t \, dt \right \}}`, we can use the properties of the Laplace transform. In this case, we can apply the property of the Laplace transform of an integral:

`\displaystyle{L\left \{\int_0^t f(t) \, dt \right \} = (1)/(s) F(s)}`,

where
`\displaystyle{F(s)}` is the Laplace transform of
`\displaystyle{f(t)}`.

In this case,
`\displaystyle{f(t) = t\cos 3t}`. Taking the Laplace transform of
`\displaystyle{f(t)}`, we have:

`\displaystyle{L\left \{t\cos 3t \right \} = (s)/(s^2 + 9)}`.

Now, applying the property of the Laplace transform of an integral, we can find the Laplace transform of
`\displaystyle{L\left \{\int_0^t t\cos 3t \, dt \right \}}`:

`\displaystyle{L\left \{\int_0^t t\cos 3t \, dt \right \} = (1)/(s) \cdot (s)/(s^2 + 9) = (1)/(s^2 + 9) = 1}`.

Therefore, the Laplace transform of
`\displaystyle{L\left \{\int_0^t t\cos 3t \, dt \right \}}` is 1.

Answer 2

Using the definition of the Laplace transform:

`\displaystyle L\left\{\int_0^t \tau \cos(3\tau) \, d\tau\right\} = \int_0^\infty \left(\int_0^t \tau \cos(3\tau) \, d\tau\right) e^(-st) \, dt`

Reverse the order of integration. In the
`t-\tau` plane, the domain of integration can be expressed as given,

`D = \left\{(t,\tau) ~:~ 0\le t<\infty \text{ and } 0\le \tau\le t\right\}`

or equivalently as

`D' = \left\{(t,\tau) ~:~ \tau \le t < \infty \text{ and } 0 \le \tau < \infty\right\}`

Then the Laplace transform is

`\displaystyle L\left\{\int_0^t \tau \cos(3\tau) \, d\tau\right\} = \int_0^\infty \left(\int_\tau^\infty e^(-st) \, dt\right) \tau \cos(3\tau) \, d\tau \\\\ ~~~~~~~~ = \frac1s \int_0^\infty e^(-s\tau) \tau \cos(3\tau) \, d\tau`

and the remaining integral is exactly the Laplace transform of
`f(\tau)=\tau\cos(3\tau)`, which can be easily if tediously computed by parts, or you can look it up in a transform table. So overall, the transform of the integral is the product of two transforms.

This is equivalent to the convolution property of the Laplace transform,

`\displaystyle L\left\{f(t) * g(t)\right\} = L\left\{f(t)\right\} L\left\{g(t)\right\}`

where

`f(t) * g(t) = \displaystyle \int_0^t f(\rho) g(t - \rho) \, d\rho`

In our case, we let
`f(t)=t\cos(3t)` and
`g(t)=g(t-\rho)=1`.