Solving $\int_0^{\infty} \frac{\sin^m(x)}{x^n} dx$ for $m, n \in \mathbb{Z}^+$ I saw this question and thought of the more general integral $$I(m, n) =\int_0^{\infty} \frac{\sin^m(x)}{x^n} dx$$
where $m, n$ are positive integers, and the integral converges. The integral converges when $2 \le n \le m$ or $n = 1, m$ is odd.
I tried substitution by parts, but was not able to get anywhere. Setting $u = \sin^m(x)$ and $dv = \frac{dx}{x^n}$ when $n \not= 1$, I got $$\frac{\sin^m(x)}{(1-n)x^{n-1}}\Biggl|^\infty_0-\int_0^{\infty}\frac{m\cos(x)\sin^{m-1}(x)}{(1-n)x^{n-1}} dx=\frac{m}{n-1}\int_0^{\infty}\frac{\cos(x)\sin^{m-1}(x)}{x^{n-1}} dx$$
This seems just as bad, if not worse, than the original integral. Even when I tried setting $u$ and $dv$ to something else, I could not get something that looked easier to solve.

Switching tracks completely, I will try to use Feynman's trick (in a similar way as Mark Viola did in the top answer to the linked question). We can write $$F(s) = \int_0^{\infty} \frac{\sin^m(x)}{x^n} e^{-sx}dx$$ 
Differentiating $F(s)$ with respect to $s$, $n$ times, yields $$F^{(n)}(s) = \int_0^{\infty} (-1)^n e^{-sx}\sin^m(x) dx$$
The integral can be rewritten as a rational function, but I realized that it would become extremely messy to integrate that $n$ times. Because of this, I didn't even bother finding the analytical expression of this.

I found without proof that $I(2k+1, 1) = \frac{\binom{2k}{k}\pi}{2^{2k+1}}$
How can I solve the integral, whether by substitution by parts, Feynman's trick, or some other method?

I'm not certain if this helps but I thought of deriving a reduction formula,
$$I(m,n) = \int_0^{\infty}\frac{\sin^mx}{x^n}dx$$ Using Integration by Parts,
$$I(m,n) = -\frac{1}{(n-1)x^{n-1}}\sin^mx|_0^{\infty}+\int_0^{\infty}\frac{m\sin^{m-1}x\cos x}{(n-1)x^{n-1}}dx=\int_0^{\infty}\frac{m\sin^{m-1}x\cos x}{(n-1)x^{n-1}}dx$$The first term is $0$ since for convergence, we need $n \leq m$.
Using Integration by Parts once more,
$$\frac{n-1}{m}I(m,n) = -\frac{1}{(n-2)x^{n-2}}\sin^{m-1}x\cos x|_0^{\infty}+\int_0^{\infty}\frac{(m-1)\sin^{m-2}x\cos^2x-\sin^mx}{(n-2)x^{n-2}}dx$$
$$\therefore \frac{m-1}{n}I(m,n)=\frac{m-1}{n-2}\int_0^{\infty}\frac{\sin^{m-2}x(1-\sin^2x)}{x^{n-2}}dx-\frac{1}{n-2}I(m,n-2)$$
$$\frac{m-1}{n}I(m,n) =\frac{m-1}{n-2}\int_0^{\infty}\frac{\sin^{m-2}x}{x^{n-2}}dx-\frac{m-1}{n-2}\int_0^{\infty}\frac{\sin^mx}{x^{n-2}}dx-\frac{1}{n-2}I(m,n-2)$$
$$\therefore \frac{m-1}{n}I(m,n) =\frac{m-1}{n-2}I(m-2,n-2)-\frac{m-1}{n-2}I(m,n-2)-\frac{1}{n-2}I(m,n-2)$$
$$\therefore \frac{m-1}{n}I(m,n)=\frac{m-2}{n-2}I(m-2,n-2)-\frac{m-2}{n-2}I(m,n-2)$$
This only works if $n >1$ which ensures the first step.
Hopefully, this leads to something.
