Question

Let f(t) be a function defined for all positive values of t. The Laplace Transform of f(t) is defined by the following integral, if the improper integral exists.F(s) = ∞e−st f(t) dt0 Laplace Transforms are used to solve differential equations. Find the Laplace Transform of the function.f(t) = cos at

Answer 1

We have laplace transform of any function is defined as

`L[{f(t)}]=\int_(0)^(\infty )f(t)e^(-st)dt`

thus for
`f(t)=cos(at)`

`L[cos(at)]=\int_(0)^(\infty )cos(at)e^(-st)dt`

Now integrating the given expression using the integration by parts principle we have

`\int cos(at)e^(-st)dt=cos(at)\int e^(-st)dt-\int (dcos(at))/(dt)\int (e^(-st)dt)dt\\\\=cos(at)(e^(-st))/(-s)-(\int -asin(at)(e^(-st))/(-s))\\\\-(cos(at)e^(-st))/(s)-(a)/(s)\int sin(at)e^(-st)dt\\\\=-(cos(at)e^(-st))/(s)-(a)/(s)(sin(at)\int e^(-st)dt-\int (dsin(at))/(dt)\int (e^(-st)dt)dt\\\\=-(cos(at)e^(-st))/(s)-(a)/(s)[((sin(at)e^(-st))/(-s))-\int acos(at)(e^(-st))/(-s)`

Note that we get the first integral back in the RHS expression

Thus solving we have

`I=-(cos(at)e^(-st))/(s)+(asin(at)e^(-st))/(s^(2))+(a)/(s)I`

Solving for I we get

`I=(e^(-st)(-scos(at)+asin(at)))/(s^(2)+a^(2))`

now applying the limits we have

`I=(e^(-st)(-scos(at)+asin(at)))/(s^(2)+a^(2))|_(0)^(\infty )\\\\I=(s)/(s^(2)+a^(2))(\because \lim_(t\rightarrow \infty)e^(-st)=0)`