Can we find a closed form for $$I(a)=\int_0^\infty e^{-x}\sin^a(x)dx$$ Mathematica can easily find closed form for integer $a$:
\begin{align*} I(0)&=1\\ I(1)&=1/2\\ I(2)&=2/5\\ I(3)&=3/10\\ I(4)&=24/85\\ I(5)&=3/13\\ I(6)&=144/629\\ I(7)&=63/325\\ \end{align*}
Perhaps this should be a comment instead of an answer, but...
I don't know if we can find a closed form expression using standard functions (maybe if we allow the gamma function?), but we can certainly define a recurrence relation.
After integrating this by parts twice, we get:
$$\int e^{-x}\sin^{a}(x)dx = \frac{e^{-x}sin^{(a-1)}(x)}{1+a^2} + \frac{a(a-1)}{a^2+1}\int e^{-x}\sin^{a-2}(x)dx$$
When we make it definite, the first term goes away entirely:
$$\int_{0}^{\infty}e^{-x}\sin^{a}(x)dx = \frac{a(a-1)}{a^2+1}\int_{0}^{\infty}e^{-x}\sin^{a-2}(x)dx$$
From here, a recurrence relation becomes obvious:
$$I(a) = \frac{a(a-1)}{a^2+1}I(a-2)$$
This isn't closed form, but it's a start... might be useful for further analysis.
This a partial answer, in case of $a$ integer.
$I(a)$ is the Laplace transform of $\sin^a(x)$ for the particular value $s=1$ $$L_x[\sin^{a}(x)](s)=\int_0^\infty\sin^a(x)e^{-sx}dx$$ The tables of Laplace Transforms gives: $$L_x[\sin^{2n}(x)](s)=\frac{(2n)!}{s(s^2+2^2)(s^2+4^2)(s^2+6^2)...(s^2+(2n)^2)}$$ $$L_x[\sin^{2n+1}(x)](s)=\frac{(2n+1)!}{(s^2+1^2)(s^2+3^2)(s^2+5^2)...(s^2+(2n+1)^2)}$$ With $s=1$ : $$\int_0^\infty\sin^{2n}(x)e^{-x}dx=\frac{(2n)!}{(1+2^2)(1+4^2)(1+6^2)...(1+(2n)^2)}$$ $$\int_0^\infty\sin^{2n+1}(x)e^{-x}dx=\frac{(2n+1)!}{(1+1^2)(1+3^2)(1+5^2)...(1+(2n+1)^2)}$$ From H.Bateman, "Tables of Integral Transforms", Eq.(3) & (7),p .150 , McGraw Hill, 1954.
In case of $a$ real, the closed form involves the Gauss hypergeometric function : http://www.wolframalpha.com/input/?i=integrate+sin%28x%29%5Eaexp%28-x%29dx+
