Answer
$$1$$
Work Step by Step
Use the formula
$$\int e^{a x} \sin b x d x=\frac{e^{a x}}{a^{2}+b^{2}}(a \sin b x-b \cos b x)+C$$
We get
\begin{align*} \int_{0}^{\infty} 2 e^{-\theta} \sin \theta d \theta&=\lim _{b \rightarrow \infty} \int_{0}^{b} 2 e^{-\theta} \sin \theta d \theta\\
&=\lim _{b \rightarrow \infty} 2\left[\frac{e^{-\theta}}{1+1}(-\sin \theta-\cos \theta)\right]_{0}^{b}\\
&=\lim _{b \rightarrow \infty}\left[\frac{-2(\sin b+\cos b)}{2 e^{b}}+\frac{2(\sin 0+\cos 0)}{2 e^{0}}\right]\\
&=0+\frac{2(0+1)}{2}\\
&=1
\end{align*}
