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.