Closed form for $\int_0^\infty e^{-x}\sin^a(x)dx$ 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.
