The most concise closed-form expression is this:
$$\int_{0}^{\infty} \frac{\sin^{2n+1}(x)}{x} dx = \frac{1}{2^{2n}}\binom{2n}{n}\frac{\pi}{2}$$
This is how you can conclude this:
In An Atlas of functions there is a formula of how to expand powers of sine:
$$sin^{r}(x) = \begin{cases} \displaystyle \frac{(-1)^n}{2^{r-1}} \sum_{j=0}^{n-1} (-1)^j \binom{r}{j} \cos\left[(r-2j)x\right] + \frac{1}{2^n} \binom{r}{n} &\quad \textrm{if} \quad r=2n \\ \displaystyle \frac{(-1)^n}{2^{r-1}}\sum_{j=0}^{n} (-1)^j \binom{r}{j} \sin\left[(r-2j)x\right] & \quad \textrm{if} \quad r=2n+1 \end{cases} $$
To prove this note:
\begin{align*}
(2i\sin(x))^{r} = (e^{ix}-e^{-ix})^{r} = & \sum_{j=0}^{r} \binom{r}{j} (-1)^{r-j}e^{ixj}e^{-ix(r-j)} =\sum_{j=0}^{r} \binom{r}{j} (-1)^{r-j}e^{-ix(r-2j)} \\
=& \sum_{j=0}^{r} \binom{r}{j}(-1)^{r-j}\left[\cos(x(r-2j))-i\sin(x(r-2j))\right]
\end{align*}
Now suppose that $r=2n+1$, then
\begin{align*}
2^{2n}(-1)^{n}(2i)\sin^{2n+1}(x) = &\sum_{j=0}^{2n+1} \binom{2n+1}{j}(-1)^{2n+1-j} \left[\cos(x(2n+1-2j))-i\sin(x(2n+1-2j))\right]\\
=& \sum_{j=0}^{n} \binom{2n+1}{j}(-1)^{2n+1-j} \left[\cos(x(2n+1-2j))-i\sin(x(2n+1-2j))\right]\\
+& \sum_{j=n+1}^{2n+1} \binom{2n+1}{j}(-1)^{2n+1-j}\left[\cos(x(2n+1-2j))-i\sin(x(2n+1-2j))\right]\\
=& \sum_{j=0}^{n} \binom{2n+1}{j}(-1)^{j}\left[-\cos(x(2n+1-2j))+i\sin(x(2n+1-2j))\right]\\
+&\sum_{j=n+1}^{2n+1} \binom{2n+1}{2n+1-j}(-1)^{j}\left[-\cos(x(2n+1-2j))+i\sin(x(2n+1-2j))\right]
\end{align*}
Lets $k=2n+1-j$ then
\begin{align*}
&\sum_{j=n+1}^{2n+1} \binom{2n+1}{2n+1-j}(-1)^{j}\left[-\cos(x(2n+1-2j))+i\sin(x(2n+1-2j))\right]\\
=& \sum_{k=0}^{n} \binom{2n+1}{k}(-1)^{k}\left[\cos(-x(2n+1-2k))-i\sin(-x(2n+1-2k))\right]
\end{align*}
Given that for $k=0,...,n$
\begin{align*}
\cos(-x(2n+1-2k))-\cos(x(2n+1-2k))&=&0\\
i\sin(x(2n+1-2k))-i\sin(-x(2n+1-2k))& =& 2i\sin(x(2n+1-2k))
\end{align*}
\begin{equation}
\therefore sin^{2n+1}(x) =\frac{(-1)^n}{2^{2n}}\sum_{j=0}^{n} (-1)^j \binom{2n+1}{j} \sin\left[(2n+1-2j)x\right] \tag{1}
\end{equation}
From complex analysis we know that:
\begin{equation}
\int_{0}^{\infty} \frac{\sin(x)}{x} = \frac{\pi}{2} \tag{2}
\end{equation}
From combinatorics we kwnow that that:
\begin{equation} \sum _{j=0}^{k} (-1)^j\binom{m}{j} = (-1)^k\binom{m-1}{k} \tag{3}
\end{equation}
Putting all together:
\begin{align*}
\int_{0}^{\infty} \frac{\sin^{2n+1}(x)}{x} dx =& \int_{0}^{\infty} \frac{(-1)^n}{2^{2n}}\sum_{j=0}^{n} (-1)^j \binom{2n+1}{j} \frac{\sin\left[(2n+1-2j)x\right]}{x} dx \quad \textrm{from (1)}\\
=& \frac{(-1)^n}{2^{2n}}\sum_{j=0}^{n} (-1)^j \binom{2n+1}{j}\int_{0}^{\infty} \frac{\sin\left[(2n+1-2j)x\right]}{x} dx
\end{align*}
If $u= (2n+1-2j)x $ then $ \displaystyle dx = \frac{1}{2n+1-2j} du$
\begin{align*}
\int_{0}^{\infty} \frac{\sin^{2n+1}(x)}{x} dx =& \frac{(-1)^n}{2^{2n}}\sum_{j=0}^{n} (-1)^j \binom{2n+1}{j}\int_{0}^{\infty} \frac{\sin\left[(2n+1-2j)x\right]}{x} dx\\
=& \frac{(-1)^n}{2^{2n}}\sum_{j=0}^{n} (-1)^j \binom{2n+1}{j}\int_{0}^{\infty} \frac{\sin(u)}{u} du\\
=& \frac{(-1)^n}{2^{2n}}\sum_{j=0}^{n} (-1)^j \binom{2n+1}{j} \frac{\pi}{2} \quad \textrm{from (2)}\\
=& \frac{(-1)^n}{2^{2n}} (-1)^{n} \binom{2n}{n}\frac{\pi}{2} \quad \textrm{from (3)}\\
=& \frac{1}{2^{2n}}\binom{2n}{n}\frac{\pi}{2}
\end{align*}
