6. Laplace Transforms of Integrals

We first saw the following properties in the Table of Laplace Transforms.

1. If `G(s)= Lap{g(t)}`, then `Lap{int_0^tg(t)dt}=(G(s))/s`.

2. For the general integral, if

`[intg(t)dt]_(t=0)`

is the value of the integral when `t=0`, then:

`Lap{intg(t)dt}` `=(G(s))/s+1/s[intg(t)dt]_(t=0)`

Examples

Use the above information and the Table of Laplace Transforms to find the Laplace transforms of the following integrals:

(a) `int_0^tcos\ at\ dt`

Answer

In this example, g(t) = cos at and from the Table of Laplace Transforms, we have:

`G(s)= Lap{cosat}` `=s/((s^2+a^2))`

Now, applying the first rule above, we have:

`Lap{int_0^tcosat\ dt}=1/sxxs/(s^2+a^2)`

`=1/(s^2+a^2)`

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(b) `int_0^te^(at)cos\ bt\ dt`

Answer

This is similar to example (a). We find the transform of the function g(t) = eatcos bt, then divide by s, since we are finding the Laplace transform of the integral of g(t) evaluated from 0 to t.

`Lap{int_0^te^(at)cosbt\ dt}=1/sxx(s-a)/((s-a)^2+b^2)`

`=(s-a)/(s((s-a)^2+b^2)`

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(c) `int_0^t te^(-3t) dt`

Answer

This follows the same process as examples (a) and (b).

Find the Laplace transform of the function `g(t)=te^(-3t)` then divide by s.

`Lap{int_0^t te^(-3t)dt}=1/sxx1/((s+3)^2)`

`=1/(s(s+3)^2)`

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(d) `int_0^tsin\ at\ cos\ at\ dt`

Answer

Recall from the Double Angle Formula that

`sin 2α = 2\ sin α\ cos α`

We can use this to re-express our integrand (the part we are integrating):

`sin at\ cos at=1/2 sin 2at`

So the Laplace Transform of the integral becomes:

`Lap{int_0^t\ sin at\ cos at\ dt}=1/2 Lap{int_0^t\ sin 2at\ dt}`

`=1/2(2a)/(s(s^2+4a^2))`

`=a/(s(s^2+4a^2))`