Right so, here is what i would like to find out:

L($\displaystyle \ e^{ax}$cosbx) , where L(f(x)) means the Laplace Transform of that function; i is the imaginary unit.

That would equal to $\displaystyle \int_{0}^{\infty}\ e^{ax}*\ e^{-px}*cosbx\,dx$ = $\displaystyle \int_{0}^{\infty}\ e^{x(a-p)}*\frac {\ e^{ibx}+\ e^{-ibx}}{2i}\,dx$ = $\displaystyle \frac {1}{2i}*(\int_{0}^{\infty}\ e^{x(a-p+ib)}\,dx + \int_{0}^{\infty}\ e^{x(a-p-ib)}\,dx )$ = $\displaystyle \frac {1}{2i}*(\frac {\ e^{x(a-p+ib)}}{a-p+ib)}$ (taken from 0 to infinity) $\displaystyle + \frac {\ e^{x(a-p-ib)}}{a-p-ib}$ (taken from 0 to infinity) ) = $\displaystyle \frac {1}{2i}*(\frac {\ e^{-x(p-a-ib)}}{a-p+ib}$ (taken from 0 to infinity) $\displaystyle + \frac {\ e^{-x(p-a+ib)}}{a-p-ib}$ (taken from 0 to infinity)) = $\displaystyle \frac {1}{2i}*(-\frac {1}{a-p+ib} -\frac {1}{a-p-ib}) = \frac {1}{2i}*(\frac {1}{p-a-ib} + \frac {1}{p-a+ib}) = \frac {1}{2i}*\frac {2(p-a)}{\((p-a)^2 + \(b^2}$ = $\displaystyle \frac {p-a}{i}*\frac {1}{\((p-a)^2 +\(b^2} $

But the problem is that it says the Laplace transform for that function is supposed to be $\displaystyle \frac {p-b}{\((p-b)^2+\(b^2}$

With that having been said and since i put up the way i solved it, i have the following question:

1) Where did i go wrong in the way i solved it?

2) What does 'p' represent in the Laplace transform ? ( i realise i should know that, but i do not and would be thankful if someone told me what it represents)

3) Why, when you calculate the integral here $\displaystyle \frac {\ e^{-x(p-a-ib)}}{a-p+ib}$ (taken from 0 to infinity) $\displaystyle + \frac {\ e^{-x(p-a+ib)}}{a-p-ib}$ (taken from 0 to infinity) 0, do you put the mnus sign in front of the paranthesis, thus changing the signs of the variables and not just leave it $\displaystyle \frac {\ e^{x(-p+a+ib)}}{a-p+ib}$ (taken from 0 to infinity) $\displaystyle + \frac {\ e^{x(-p+a-ib)}}{a-p-ib}$ (taken from 0 to infinity) 0? Can the p not be negative? If you were to not move the minus sign out of the paranthesis, it would lead to and infinity case?

4) I did not know how to make the sign (taken from 0 to infinity) in LaTex. Could someone post it up in a post, please?

I would greatly appreciate help on this matter, thank you in advance!