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**Jhevon** ok. now, you do realize that the $\displaystyle x$ in what you just worked out is a constant right? (we were integrating with respect to y). this constant represents $\displaystyle a - bi$, a general complex number mentioned by Krizalid in post #36.

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

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

of course, the integral $\displaystyle \int_0^\infty e^{-x(a - bi)}~dx$ was a LOT easier to compute than the by parts method. because $\displaystyle a - bi$ 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