Find the Laplace transform of f(t)=cos at, where a is a real constant.

Answer: F(s)=s/(s^2+a^2), s>0

Find the Laplace transform of f(t)=cos at, where a is a real constant.

Answer: F(s)=s/(s^2+a^2), s>0

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Let's Lf(t) be the Laplace transform of cos at (this is like cos-Furie trabsform of the fucntion e
^{-st}).

Lf(t) = ∫_{0},^{∞} cos at e
^{-st}dt (1)

If you integrate twice in parts, the same integral (1) will come out again. You can make sure that this integral satsfies to the equation

Lf(t)= (1/s) - (a^{2}/s^{2}) Lf(t) (2)

Solve the eqautio for Lf(t), and you will see that

Lf(t) = s/(s^{2} + a^{2})

If you will have problems with integration, let me know.

Hello Sun,

The Laplace transform of a function f(t) defined for t≥0 is

£{f(t)} = ∫_{0}^{∞} e^{-st} f(t) dt,

provided of course that the above converges.

In your case, f(t) = cos(at), so you need to evaluate

£{cos(at)} = ∫_{0}^{∞} e^{-st} cos(at) dt.

This is an integration-by-parts (twice) type integral, where you name the integral I, for instance, integrate by parts twice, and find I in the resulting second integration, whereafter you solve algebraically for I.

The restriction on s (namely, s>0) comes from being able to evaluate to zero the limits of integration for the exponential function.

I hope this helps. Someone else will probably come by and solve the integration for you later.

Regards,

Hassan H.

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## Comments

Just a minor side note to what Grigori S. has contributed---his equation (2) above is not the only possible form of the algebraic relation you get, depending on your choice of which function to integrate, and which to differentiate, in your integrations-by-parts.

Regards,

Hassan H.