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. $$\int_{0}^{\infty}\frac{\sin^{2n+1}\left(x\right)}{x}dx=\frac{\pi\left(2n\right)!}{2^{2n+1}\left(n!\right)^{2}}$$
