Solving an integral with a complex component I am trying to find
$$
\int_{0}^{\infty} e^{-at} \sin{bt} \ dt
$$ 
where $a>0$ and $b$ is constant.
I'm familiar with Euler's formula, and also derived the identity
$$
\int e^{(-a+ib)t} \ dt = \frac{(-a-ib)e^{(-a+ib)t}}{a^{2}+b^{2}}
$$
which looked hopeful. 
Then applying Euler's formula in the first step,
\begin{align*}
\int_{0}^{\infty} e^{-at} \sin{bt} \ dt &= \frac{1}{2i}\int_{0}^{\infty} e^{-at}\big(e^{ibt} - e^{-ibt}\big) \ dt \\
&= \frac{1}{2i} \int_{0}^{\infty}\big(e^{(-a+ib)t} - e^{(-a-ib)t}\big) \ dt
\end{align*}
... but now I'm unsure how to continue. Using my derived identity above the first integral term is simple, but the second term is causing me trouble. I tried applying a complementary identity involving a subtraction instead of an addition to the second term, however it did not appear to be simplifiable. 
Any help or hints on how to proceed from here would be greatly appreciated. 
 A: Assuming $a>0$,
$$I(a,b)=\int_{0}^{+\infty}e^{-at}\sin(bt)\,dt = \text{Im}\int_{0}^{+\infty}e^{-(a-ib)t}\,dt $$
hence
$$ I(a,b) = \text{Im}\left(\frac{1}{a-ib}\right) = \text{Im}\left(\frac{a+ib}{a^2+b^2}\right) = \color{red}{\frac{b}{a^2+b^2}}.$$
A: $\frac 1{2i}\int_{0}^{\infty}\big(e^{(-a+ib)t} - e^{(-a-ib)t}\big) \ dt\\
\int e^{(-a+ib)t} \ dt = \frac{(-a-ib)e^{(-a+ib)t}}{a^{2}+b^{2}}\\
\int e^{(-a-ib)t} \ dt = \frac{(-a+ib)e^{(-a-ib)t}}{a^{2}+b^{2}}$
$\dfrac 1{2i} \dfrac {-a(e^{(-a+ib)t}-e^{(-a-ib)t}) -ib(e^{(-a+ib)t}+e^{(-a-ib)t})}{a^2 + b^2}\\
\dfrac 1{2i} \dfrac {-a (e^{-at}) (2i \sin bt) - ib(e^{-at})(2\cos bt)}{a^2 + b^2}
\\
\dfrac {-e^{-at}(a\sin bt + b\cos bt)}{a^2 + b^2}\\$
That takes care of the integration.
Apply the limits.
$\dfrac {b}{a^2 + b^2}\\$
A: You can also use parts twice, letting $$I = \int_0^\infty e^{-at}sin(bt)dt$$ with $u=e^{-at}$ and $dv=sin(bt)dt$ we have $$I=\frac{-1}{b}e^{-at}cos(bt)|_0^\infty - \frac{a}{b}\int_0^\infty e^{-at}cos(bt)dt. $$  The first term is $\frac{-1}{b}$ unless $a=0$ but in that case the integral is divergent.  Using parts once more with $u=e^{-at}$ and $dv=cos(bt)dt$ we have $$\frac{1}{b}-\frac{a}{b}[\frac{1}{b}sin(bt)e^{-at}|_0^\infty+\frac{a}{b}\int_0^\infty e^{-at}sin(bt)dt].$$  The second term is zero and we realize we have our original integral, I, back so we have, $$I = \frac{1}{b}-\frac{a^2}{b^2}I$$  Our integration  problem has been turned into a algebra problem, solving for I gives, $$I = \frac{b}{a^2+b^2}.$$
