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Math Help - Derivative Tutorial

  1. #46
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    OHh...now I see sort of where this is coming from...I know Euler's formula to be e^{ix}=\cos(x)+i\sin(x)...so then obviously the real part (i.e. Re[e^{ix}]) is cos(x) and the imaginary part(i.e. Im[e^{ix}]) is sin(x)...ok so I understand that and the other part about the identity makes sense since \int_0^{\infty}e^{-xy}dy=\frac{-e^{-xy}}{x}\bigg|_0^{\infty}=0-\frac{-1}{x}=\frac{1}{x}...so now the cartesian form of this would be something to the affect that I use the complex argument or something on sin(x)? Could you explain from here?
    ok. now, you do realize that the x in what you just worked out is a constant right? (we were integrating with respect to y). this constant represents a - bi, a general complex number mentioned by Krizalid in post #36.

    so that, your answer to this integral is in fact \frac 1{a - bi}

    now if we rationalize this, we get \frac {a + bi}{a^2 + b^2}. The imaginary part of this complex number is \frac {bi}{a^2 + b^2}, so that \int_0^\infty e^{-ax} \sin bx~dx = \Im \left \{ \int_0^\infty e^{-x(a - bi)} ~dx \right \} = \frac b{a^2 + b^2} as desired.

    of course, the integral \int_0^\infty e^{-x(a - bi)}~dx was a LOT easier to compute than the by parts method. because a - bi is simply a constant (complex constant, but a constant nonetheless)

    one of the nice things about complex analysis is that a lot of the things in real calculus generalizes well with things in complex calculus, including how we integrate e raised to some constant times x, in general
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  2. #47
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Jhevon View Post
    ok. now, you do realize that the x in what you just worked out is a constant right? (we were integrating with respect to y). this constant represents a - bi, a general complex number mentioned by Krizalid in post #36.

    so that, your answer to this integral is in fact \frac 1{a - bi}

    now if we rationalize this, we get \frac {a + bi}{a^2 + b^2}. The imaginary part of this complex number is \frac {bi}{a^2 + b^2}, so that \int_0^\infty e^{-ax} \sin bx~dx = \Im \left \{ \int_0^\infty e^{-x(a - bi)} ~dx \right \} = \frac b{a^2 + b^2} as desired.

    of course, the integral \int_0^\infty e^{-x(a - bi)}~dx was a LOT easier to compute than the by parts method. because a - bi is simply a constant (complex constant, but a constant nonetheless)

    one of the nice things about complex analysis is that a lot of the things in real calculus generalizes well with things in complex calculus, including how we integrate e raised to some constant times x, in general
    Ok I have some questions...first off yes I do realize it was a constant...it represents a+0i in this case yeah? Or does it always represent the Im part also?

    So I get that \int_0^{\infty}e^{(a-bi)y}dy=\frac{1}{a-bi}

    Now what I dont understand is hwo it made the \sin(bx) dissapear

    Next thing is I will assume since we integrated in repect to the imaginary part we want the imaginary part that is why we took \frac{bi}{a^2+b^2}.....but why did we discard the bi in lieu of just b?

    Thanks so much for your effort...I truly appreciate it..
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  3. #48
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Ok I have some questions...first off yes I do realize it was a constant...it represents a+0i in this case yeah? Or does it always represent the Im part also?

    So I get that \int_0^{\infty}e^{(a-bi)y}dy=\frac{1}{a-bi}

    Now what I dont understand is hwo it made the \sin(bx) dissapear

    Next thing is I will assume since we integrated in repect to the imaginary part we want the imaginary part that is why we took \frac{bi}{a^2+b^2}.....but why did we discard the bi in lieu of just b?

    Thanks so much for your effort...I truly appreciate it..
    the sin(bx) did not disappear, it is there (except here, we made the x into y, but it does not matter).

    \int_0^\infty e^{-x(a - bi)}~dx = \int_0^\infty e^{-ax}(cos(bx) + i \sin (bx))~dx =  \int_0^\infty e^{-ax} \cos (bx)~dx + i \int_0^\infty e^{-ax} \sin (bx)~dx .....by Euler's formula

    the last integral is what we wanted. of course, we would disregard the i, that is why we just took the \frac b{a^2 + b^2} from the \frac {bi}{a^2 + b^2}, we did not care about the i part, only its coefficient, which happens to be the integral we want to find

    if we wanted to find \int_0^\infty e^{-ax} \cos bx~dx, we would take the real part of the answer, and so the answer would be \frac a{a^2 + b^2}
    Last edited by Jhevon; April 24th 2008 at 11:25 PM.
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