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Math Help - Laplace transform

  1. #1
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    Laplace transform

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

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

    That would equal to \int_{0}^{\infty}\ e^{ax}*\ e^{-px}*cosbx\,dx = \int_{0}^{\infty}\ e^{x(a-p)}*\frac {\ e^{ibx}+\ e^{-ibx}}{2i}\,dx = \frac {1}{2i}*(\int_{0}^{\infty}\ e^{x(a-p+ib)}\,dx + \int_{0}^{\infty}\ e^{x(a-p-ib)}\,dx ) = \frac {1}{2i}*(\frac {\ e^{x(a-p+ib)}}{a-p+ib)} (taken from 0 to infinity) + \frac {\ e^{x(a-p-ib)}}{a-p-ib} (taken from 0 to infinity) ) = \frac {1}{2i}*(\frac {\ e^{-x(p-a-ib)}}{a-p+ib} (taken from 0 to infinity) + \frac {\ e^{-x(p-a+ib)}}{a-p-ib} (taken from 0 to infinity)) = \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} = \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 \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 \frac {\ e^{-x(p-a-ib)}}{a-p+ib} (taken from 0 to infinity) + \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 \frac {\ e^{x(-p+a+ib)}}{a-p+ib} (taken from 0 to infinity) + \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!
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  2. #2
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    First note the complex form of cosine does not have an i in the denominator that is sine.

    \displaystyle \int_{0}^{\infty}\ e^{ax} e^{-px}\cos(bx)dx=

    \displaystyle \int_{0}^{\infty}\ e^{ax} e^{-px}\left( \frac{e^{ibx}+e^{-ibx}}{2}\right)dx=
    \displaystyle \frac{1}{2} \int_{0}^{\infty} e^{-x(p-a-ib)}+e^{-x(p-a+ib)}dx=\frac{1}{2}\left( \frac{-e^{-x(p-a-ib)}}{[(p-a)-ib]} \bigg|_{0}^{\infty}+ \frac{-e^{-x(p-a+ib)}}{[(p-a)+ib]} \bigg|_{0}^{\infty}\right)

    \displaystyle \frac{1}{2}\left( \frac{1}{[(p-a)-ib]}+\frac{1}{[(p-a)+ib]}\right) =\frac{1}{2}\left( \frac{(p-a)+ib+(p-a)-ib}{[(p-a)-ib][(p-a)+ib]}\right)=\frac{(p-a)}{(p-a)^2+b^2}

    As for p is the Laplace transform it is a complex number with is real part such that the integral above converges so for the above example if \text{Re}(p) > a then as x \to \infty the integral converges, otherwise it would diverge.
    Last edited by TheEmptySet; December 7th 2010 at 11:40 AM. Reason: Correct an error
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  3. #3
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    From the solution that it came to, i draw 2 conclusion:

    1) The notes i have taken in class for \displaystyle cos(x)=\frac {\ e^{ix}+\ e^{-ix}}{2i} was wrong, the denominator being only 2.

    2) The solution i thought was supposed to be the correct one, namely \displaystyle \frac {p-b}{(p-b)^2 + a^2} , was also most probably wrongfully heard by me, as the correct solution is staring me in the face.

    Now, for the part about p. From what i understood, since p by its nature is a number put there to make the integral converge, if we take the \displaystyle \frac{1}{2} \int_{0}^{\infty} e^{-x(p-a-ib)} as a Laplace transform on it's own, which it is, it HAS to be convergent, and the only way for it to be convergent (not diverge towards infinity) is by taking the minus sign out of the paranthesis, which in term will make Re(p) > a, and the integral result as a finite solution, which is what the Laplace Transform must be, otherwise there would be no point in transforming a function towards another domain because if it diverges you cannot transport it back to its original domain and find the required solution.

    Am i correct on my judgement?

    Also, how do we know 'a' isn't actually a number smaller than Re(p), which means Re(p) > a without putting the minus sign out of the paranthesis? I am assuming you were refering to -p being smaller than a? If not then where did you deduce that Re(p)>a if not from p being greater than -a?
    Last edited by kamykazee; December 7th 2010 at 12:18 PM.
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