I tried differentiating under the integral sign which led me to: $$f'\left(y\right)=-\int_{0}^{\infty}e^{-xy}\sin^{2n+1}\left(x\right)dx\ =\frac{-\left(2n+1\right)!}{\prod_{k=0}^{n}\left(y^{2}+\left(2k+1\right)^{2}\right)}$$ where $$f\left(y\right)=\int_{0}^{\infty}e^{-xy}\frac{\sin^{2n+1}\left(x\right)}{x}dx$$ And now I'm stuck trying to evaluate $$-\int_{b}^{y\ }\frac{\left(2n+1\right)!}{\prod_{k=0}^{n}\left(t^{2}+\left(2k+1\right)^{2}\right)}dt=f\left(y\right)-f\left(b\right)$$ which, once I let $b$ approach $\infty$ and prove that I can set $y=0$ in the resulting expression, will lead me to the answer. I know the integral in $t$ is technically doable using partial fractions but I can't see how I can get an explicit expression in terms of $n$ (which I know exists) out of this approach.  Here's the answer. $$\int_{0}^{\infty}\frac{\sin^{2n+1}\left(x\right)}{x}dx=\frac{\pi\left(2n\right)!}{2^{2n+1}\left(n!\right)^{2}}$$
Here is a fun integral I am trying to evaluate:  $$\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \ dx=\frac{\pi \binom{2n}{n}}{2^{2n+1}}.$$  I thought about integrating by parts $2n$ times and then using the binomial theorem for $\sin(x)$, that is, using $\dfrac{e^{ix}-e^{-ix}}{2i}$ form in the binomial series.  But, I am having a rough time getting it set up correctly. Then, again, there is probably a better approach.   $$\frac{1}{(2n)!}\int_{0}^{\infty}\frac{1}{(2i)^{2n}}\sum_{k=0}^{n}(-1)^{2n+1-k}\binom{2n}{k}\frac{d^{2n}}{dx^{2n}}(e^{i(2k-2n-1)x})\frac{dx}{x^{1-2n}}$$  or something like that. I doubt if that is anywhere close, but is my initial idea of using the binomial series for sin valid or is there a better way?.  Thanks everyone.
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