Let $I(m,n):=\displaystyle \int_{0}^{\infty} \frac{\sin ^{m} x}{x^{n}} d x$, where $0<n\leq m.$
A month ago, I had found,in my essay, that when $m$ and $n$ are of the same parity and $2\leq n\leq m,$ $$\boxed{\int_0^{\infty} \frac{\sin^{m}x}{x^{n}}dx=\frac{(-1)^{\frac{m-n}{2}} \pi}{2^{m}(n-1) !} \sum_{k=0}^{\left\lfloor\frac{m-1}{2}\right\rfloor}(-1)^{k}\left(\begin{array}{l} m \\ k \end{array}\right)(m-2 k)^{n-1}}, $$
In other words, $I(m,n) =\dfrac{a\pi}{b}$ for some natural numbers $a$ and $b$.
However, when $m$ and $n$ are of different parity, $$I(m,n) =\dfrac{a}{b}ln(\dfrac{c}{d})$$ for some natural numbers $a, b, c $ and $d$.
Then I started to investigate whether there is a formula for $I(m,n)$ of different parity.
Started from easy, I am going to find, by Frullani’s Integral Theorem), a formula for, $$I(2n+1,2)=\int_{0}^{\infty} \frac{\sin ^{2n+1} x}{x^{2}} d x =\frac{2 n+1}{2^{2 n}(-1)^{n-1}} \sum_{k=0}^{n-1}(-1)^{k}\left(\begin{array}{c} 2 n-1 \\ k \end{array}\right) \ln \left|\frac{2 n+1-2 k}{2 n-3-2 k}\right|. $$ Proof: $$ \begin{aligned} I(2 n+1,2) &=\int_{0}^{\infty} \sin ^{2 n+1} x d\left(-\frac{1}{x}\right) \\ &\stackrel{IBP}{=} \int_{0}^{\infty} \frac{(2 n+1) \sin ^{2 n} x \cos x}{x} d x \\ &=(2 n+1) \int_{0}^{\infty} \frac{\sin 2 x}{2 x} \cdot \sin ^{2 n-1} x d x \end{aligned} $$ Expressing $\sin^{2n-1}x$ as a linear combination of $\sin(2n-1-2k)$ yields $$\begin{array}{l} \\ \displaystyle \quad I(2n+1,2)\\ \displaystyle =\frac{2 n+1}{2} \int_{0}^{\infty} \frac{\sin 2 x}{x} \left[\frac{1}{2^{2 n-2}(-1)^{n-1}} \sum_{k=0}^{n-1}(-1)^{k} \left(\begin{array}{c} 2 n-1 \\ k \end{array}\right)\sin (2 n-1-2 k) x \right]dx\\ \displaystyle =\frac{2 n+1}{2^{2 n-1}(-1)^{n-1}} \sum_{k=0}^{n-1}(-1)^{k}\left(\begin{array}{c} 2 n-1 \\ k \end{array}\right)\int_{0}^{\infty} \frac{\sin 2 x \sin (2 n-1-2 k)x}{x} d x \\ =\displaystyle \frac{2 n+1}{2^{2 n}(-1)^{n-1}} \frac{n-1}{k=0}^{n-1}(-1)^{k}\left(\begin{array}{c} 2 n-1 \\ k \end{array}\right) \int_{0}^{\infty} \frac{\cos (2 n-3-2 k) x-\cos (2 n+1-2k)x}{x} dx\\ \displaystyle =\frac{2 n+1}{2^{2 n}(-1)^{n-1}} \sum_{k=0}^{n-1}(-1)^{k}\left(\begin{array}{c} 2 n-1 \\ k \end{array}\right) \ln \left|\frac{2 n+1-2 k}{2 n-3-2 k}\right|\quad \blacksquare \end{array}$$(The last step using Frullani’s Integral Theorem)
For examples: $$ \begin{aligned} \int_{0}^{\infty} \frac{\sin ^{5} x}{x^{2}} d x &=\frac{-5}{16}\left[\left(\begin{array}{c} 3 \\ 0 \end{array}\right) \ln 5-\left(\begin{array}{l} 3 \\ 1 \end{array}\right) \ln 3\right]=\frac{5}{16} \ln \frac{27}{5} \end{aligned} $$ $$ \int_{0}^{\infty} \frac{\sin ^{7} x}{x^{2}} d x=\frac{7}{64}\left[\left(\begin{array}{l} 5 \\ 0 \end{array}\right) \ln \left(\frac{7}{3}\right)-\left(\begin{array}{c} 5 \\ 1 \end{array}\right) \ln \left(\frac{5}{1}\right)+\left(\begin{array}{l} 5 \\ 2 \end{array}\right) \ln \left(\frac{3}{1}\right)\right]= \frac{7}{64} \ln \left(\frac{137781}{3125}\right) $$ Eureka! I succeeded to find a formula for $I(2n+1,2)$. I believe that we can find a reduction formula and use Induction to prove that when $m$ and $n$ are of different parity, $$I(m,n) =\dfrac{a}{b}\ln(\dfrac{c}{d})$$ However, it is difficult to find a general formula for $I(m,n)$ of different parity.
Please give me opinions to go further. Thank you very much!
