How to find the closed form of the integral$\int_{0}^{\infty}f(x)\frac{\sin^nx}{x^m}dx$

Where f(x) is an even function,and also is a periodic functions,the period is $\pi$,and $n,m\in\Bbb N$,and n+m is odd number. When n+m is even, I already have a solution.But when n+m is odd, I don't know how to solve it.I'd appreciate it if someone could help me with it.

When n+m is even,my answer is as follows:

If f (x) is an even function, and the period is $\pi$,we have: $$\int_{0}^\infty f(x)\frac{\sin^nx}{x^m}dx=\int_{0}^\frac{\pi}{2}f(x)g_m(x)\sin^nxdx \qquad (1)$$

$n,m\in\Bbb N$, Where the n+m is an even,and $g_m(x)$ in (1) is as follows: $$g_m(x)=\begin{cases}\frac{(-1)^{m-1}}{(m-1)!}\frac{d^{m-1}}{dx^{m-1}}\left(\csc x\right),& \text{for n is odd and}\\[2ex] \frac{(-1)^{m-1}}{(m-1)!}\frac{d^{m-1}}{dx^{m-1}}\left(\cot x\right),& \text{ for n is even .} \end{cases}$$ —————————————————————————————————————————————————— Proof: \begin{align} \int_{0}^\infty f(x)\frac{\sin^nx}{x^m}dx&=\sum_{k=0}^\infty\int_{k\pi}^{(2k+1)\frac{\pi}{2}}f(x)\left(\frac{\sin ^nx}{x^m}\right)dx+\sum_{k=1}^\infty\int_{(2k-1)\frac{\pi}{2}}^{k\pi}f(x)\left(\frac{\sin^n x}{x^m}\right)dx\\ &=\sum_{k=0}^\infty\int_{0}^{\frac{\pi}{2}}f(x+k\pi)\left(\frac{\sin^n (x+k\pi)}{(x+k\pi)^m}\right)dx+\sum_{k=1}^\infty\int_{-\frac{\pi}{2}}^{0}f(x+k\pi)\left(\frac{\sin^n (x+k\pi)}{(x+k\pi)^m}\right)dx\\ &=\sum_{k=0}^\infty(-1)^{nk}\int_{0}^{\frac{\pi}{2}}f(x)\left(\frac{\sin^n x}{(x+k\pi)^m}\right)dx+\sum_{k=1}^\infty(-1)^{nk+n+m}\int_{0}^{\frac{\pi}{2}}f(-x)\left(\frac{\sin^n x}{(x-k\pi)^m}\right)dx\\ &=\int_{0}^{\frac{\pi}{2}}f(x)\sin^nx\left(\frac{1}{x^m}+\sum_{k=1}^\infty(-1)^{nk}\left[\frac{1}{(x+k\pi)^m}+\frac{(-1)^{n+m}}{(x-k\pi)^m}\right]\right)dx\\ &=\int_{0}^{\frac{\pi}{2}}f(x)\sin^nxg_m(x)dx \end{align} When n+m is an even,and we know by the Fourier series \begin{align} \csc x&=\frac{1}{x}+\sum_{k=1}^\infty(-1)^k\left(\frac{1}{x+k\pi}+\frac{1}{x-k\pi}\right)\\ \end{align} and \begin{align} \cot x&=\frac{1}{x}+\sum_{k=1}^\infty\left(\frac{1}{x+k\pi}+\frac{1}{x-k\pi}\right) \end{align} Take the m-1 order derivative,thus we obtain $g_m(x)$. —————————————————————————————————————————————————— Example: \begin{align} (1.)\qquad\int_{0}^{\infty}\frac{\sin^3x}{x}dx&=\int_{0}^{\frac{\pi}{2}}\sin^2xg_1(x)\sin xdx\\ &=\int_{0}^{\frac{\pi}{2}}\sin^2x\frac{1}{\sin x}\sin xdx\\ &=\int_{0}^{\frac{\pi}{2}}\sin^2xdx\\ &=\frac{\pi}{4}\\ \end{align} \begin{align} (2.) \int_{0}^{\infty}(1+\cos^2x)\frac{\sin^2x}{x^2}dx &=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)g_2(x)\sin^2xdx\\ &=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)\left(-\frac{d}{dx}\cot x\right)\sin^2xdx\\ &=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)\left(\frac{1}{\sin^2x}\right)\sin^2xdx\\ &=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)dx\\ &=\frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}\\ \end{align} \begin{align} (3.) \int_{0}^{\infty}\frac{1}{(1+\cos^2x)}\frac{\sin^3x}{x^3}dx &=\int_{0}^{\frac{\pi}{2}}\frac{\sin^3x}{(1+\cos^2x)}g_3(x)dx\\ &=\int_{0}^{\frac{\pi}{2}}\frac{\sin^3x}{(1+\cos^2x)}\left(\frac{1}{2}\frac{d^2}{dx^2}(\csc x)\right)dx\\ &=\int_{0}^{\frac{\pi}{2}}\frac{\sin^3x}{(1+\cos^2x)}\frac{(1+\cos^2x)}{2\sin^3x}dx\\ &=\int_{0}^{\frac{\pi}{2}}\frac{1}{2}dx=\frac{\pi}{4}\\ \end{align}

Solution 1:

There is an excellent result related with this integral, enjoy!

$$I=\int_{0}^{\infty }x^{p}\ \left ( \frac{\sin(x)}{x} \right )^ndx\ \ \ \ \ \ , n=1,2,3,\dots , \ \ \ 0\geq p\geq -1,\\ \\ \\ I=\frac{\pi }{2(2i)^{n}\Gamma (n-p)}\sum_{m=0}^{n }(-1)^{n-m}\frac{n!}{m!(n-m)!}\left | n-2m \right |^{n-p-1}\left ( \frac{1}{\sin(\frac{n-p+1}{2})\pi }-\frac{\operatorname{sgn}(n-2m)}{\sin(\frac{n-p}{2})\pi }i \right ).$$