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?

• The most straightforward method I'm aware of is actually doing the messy algebra. For small values this is doable and one can (more or less) generalize it to obtain a general formiula; anyway, the formula is arguably not pretty... (see equation $(37)$). I've sketched the general approach for $n=m$ here. Commented Apr 30, 2020 at 13:49
• That formula was exactly what I was looking for! Thank you. Commented Apr 30, 2020 at 15:18

In this essay, I had found a formula for the integrals with powers of the same parity.

$$\boxed{\int_0^{\infty} \frac{\sin^{m}x}{x^{n}}dx=\frac{(-1)^{\frac{m-n}{2}} \pi}{2^{m}(n-1) !} \sum_{k=0}^{\left\lfloor\frac{m-1}{2}\right\rfloor}(-1)^{k}\left(\begin{array}{l} m \\ k \end{array}\right)(m-2 k)^{n-1}},$$

$$\text{where }m$$ and $$n$$ are of the same parity and $$2\leq n\leq m.$$

In particular, when $$m=n$$, we had completed a formula for:

$$\boxed{\int_{0}^{\infty} \frac{\sin ^{n} x}{x^{n}}=\frac{\pi}{2^{n}(n-1) !} \sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}(-1)^{k}\left(\begin{array}{c} n \\ k \end{array}\right)(n-2 k)^{n-1} }$$ $$\textrm{where }n\in N.$$

First of all, I start with the integral of even powers in both numerator and denominator. Let

$$\displaystyle I(m, n):=\int \frac{\sin^{2m}x}{x^{2n}}dx,\tag*{}$$

where $$0

Using the formula below, we first express $$\sin^{2m}x$$ as a linear combination of $$\cos 2kx$$, whose high derivatives can be easily found.

$$\sin ^{2 m} x=\frac{1}{2^{2 m-1}(-1)^{m}} \sum_{k=0}^{m-1}(-1)^{k}\left(\begin{array}{c}2m\\ k\end{array}\right) \cos (2 m-2 k) x+\frac{1}{2^{2m}} \left(\begin{array}{c}2 m \\ m\end{array}\right)\quad \text{(See footnotes for proof.)}$$ Now differentiate $$\sin ^{2 m} x$$ w.r.t. $$x$$ by $$2n-1$$ times. $$\left(\sin ^{2 m} x\right)^{(2 n-1)}=\frac{1}{2^{2 m-1}(-1)^{m}} \sum_{k=0}^{m-1}(-1)^{k+1}\left(\begin{array}{c}2 m \\ k\end{array}\right)(-1)^{n-1}(2 m-2 k)^{2 n-1} \sin (2 m-2k)x$$ $$=\frac{(-1)^{m-n}}{2^{2 m-1}} \sum_{k=0}^{m-1}(-1)^{k}\left(\begin{array}{c}2 m \\ k\end{array}\right)(2 m-2 k)^{2 n-1} \sin (2 m-2 k) x$$

Applying integration by parts $$2n-1$$ times yields

$$\displaystyle I(m, n)=\int_{0}^{\infty} \frac{\left(\sin ^{2 m} x\right)^{(2 n-1)}}{(2 n-1) ! x} d x \\ \\ \displaystyle \quad \qquad =\frac{(-1)^{m-n}}{2^{2 m-1}(2 n-1) !} \sum_{k=0}^{m-1}(-1)^{k}\left(\begin{array}{c}2 m \\ k\end{array}\right)(2 m-2 k)^{2 n-1} \int_{0}^{\infty} \frac{\sin (2 m-2 k) x}{x} d x$$

Using the well-known result $$\displaystyle \int_{0}^{\infty} \frac{\sin a x}{x} d x=\frac{\pi}{2}$$ for any $$a>0,$$ we can conclude that

$$\displaystyle I(m, n) =\frac{(-1)^{m-n} \pi}{2^{2 m}(2 n-1) !} \sum_{k=0}^{m-1}(-1)^{k}\left(\begin{array}{c}2 m \\ k\end{array}\right)(2 m-2 k)^{2 n-1}.\tag*{(1)}$$

Now I come back to the integrals with odd powers $$J(m, n):=\int_{0}^{\infty} \frac{\sin ^{2 m+1} x}{x^{2 n+1}} d x$$

By the other similar formula for odd powers of sine:

$$\sin ^{2 m+1} x=\frac{1}{2^{2 m} (-1)^{m}} \sum_{k=0}^{m}(-1)^{k}\left(\begin{array}{c}2 m+1 \\ k\end{array}\right) \sin (2 m+1-2 k) x$$ We differentiate the equation w.r.t.$$x$$ by $$2n$$ times and get $$\left(\sin ^{2 m+1} x\right)^{(2 n)}=\frac{(-1)^{m-n}}{2^{2 m}} \sum_{k=0}^{m}(-1)^{k}\left(\begin{array}{c}2 m+1 \\ k\end{array}\right)(2 m+1-2 k)^{2 n}\sin (2 m+1-2 k) x$$ Applying integration by parts $$2n$$ times yields

$$\begin{array}{l}\displaystyle \int_{0}^{\infty} \frac{\sin ^{2 m+1} x}{x^{2 n+1}} d x=\int_{0}^{\infty} \frac{\left(\sin ^{2 m+1} x\right)^{(2 n)}}{(2 n) ! x} d x \end{array}$$ Plugging the $$2n^{th }$$derivative gives $$\begin{array}{l} \displaystyle J(m,n) =\frac{(-1)^{m-n}}{2^{2 m}(2 n) !} \sum_{k=0}^{m}(-1)^{k}\left(\begin{array}{c} 2 m+1 \\ k \end{array}\right)(2 m+1-2 k)^{2 n} \int_{0}^{\infty} \frac{\sin (2 m+1-2 k)}{x} \\ \\ \displaystyle \quad \qquad =\frac{(-1)^{m-n} \pi}{2^{2 m+1}(2 n) !} \sum_{k=0}^{m}(-1)^{k}\left(\begin{array}{c} 2 m+1 \\ k \end{array}\right)(2 m+1-2 k)^{2 n}\tag*{(2)} \end{array}$$

Merging the formula (1) and (2), we can now conclude in one single formula that

$$\boxed{\int_0^{\infty} \frac{\sin^{m}x}{x^{n}}dx=\frac{(-1)^{\frac{m-n}{2}} \pi}{2^{m}(n-1) !} \sum_{k=0}^{\left\lfloor\frac{m-1}{2}\right\rfloor}(-1)^{k}\left(\begin{array}{l} m \\ k \end{array}\right)(m-2 k)^{n-1}},$$ In particular, when $$m=n$$, we had completed the formula for:

$$\boxed{\int_{0}^{\infty} \frac{\sin ^{n} x}{x^{n}}=\frac{\pi}{2^{n}(n-1) !} \sum_{k=0}^{\left\lfloor\frac{n-1}{2}\right\rfloor}(-1)^{k}\left(\begin{array}{c} n \\ k \end{array}\right)(n-2 k)^{n-1} }$$ $$\textrm{where }n\in N.$$

By the way, we can easily get any results by putting values of $$n$$ in the formula. For example, Example 1

\begin{aligned} & \int_{0}^{\infty} \frac{\sin ^{7} x}{x^{3}} d x \\ =& \frac{(-1)^{\frac{7-3}{2}} \pi}{2^{7}(3-1) !} \sum_{k=0}^{3}(-1)^{k}\left(\begin{array}{l} 7 \\ k \end{array}\right)(7-2 k)^{2} \\ =& \frac{\pi}{2^{8}}\left[\left(\begin{array}{l} 7 \\ 0 \end{array}\right) 7^{2}-\left(\begin{array}{c} 7 \\ 1 \end{array}\right) 5^{2}+\left(\begin{array}{l} 7 \\ 2 \end{array}\right) 3^{2}-\left(\begin{array}{l} 7 \\ 3 \end{array}\right) 1^{2}\right] \\ =& \frac{\pi}{2^{8}}(49-175+189-35) \\ =& \frac{7 \pi}{64} \end{aligned}

Example 2 $$\begin{array}{rl} \displaystyle \displaystyle \int_{0}^{\infty} \frac{\sin ^{9} x}{x^{9}} dx=\displaystyle \frac{\pi}{2^{9} \cdot 8 !}\left[\left(\begin{array}{l} 9 \\ 0 \end{array}\right) 9^{8}-\left(\begin{array}{l} 9 \\ 1 \end{array}\right) 7^{8}+\left(\begin{array}{l} 9 \\ 2 \end{array}\right) 5^{8}-\left(\begin{array}{l} 9 \\ 3 \end{array}\right) 3^{8}+\left(\begin{array}{l} 9 \\ 4 \end{array}\right)\right] \end{array} \displaystyle =\frac{259723 \pi}{1146880}$$

Footnotes:

1. Let $$z=\cos \theta+i \sin \theta$$, then $$\displaystyle \sin x=\frac{z-\bar{z}}{2 i}$$ and

$$\displaystyle \quad \sin ^{2 n} x=\left(\frac{z-\overline{z}}{2 i}\right)^{2n}=\frac{1}{2^{n}(-1)^{n}} \sum_{k=0}^{2 n}\left(\begin{array}{c}2 n \\ k\end{array}\right) z^{2 n-k}\left(-\bar z\right)^{k}\\ \displaystyle \qquad =\frac{1}{2^{2 n}(-1)^{n}} \left[\sum_{k=0}^{n-1}(-1)^{k}\left(\begin{array}{c}2 n \\ k\end{array}\right)\left(z^{2 n-2 k}+\bar z^{2 n-2 k}\right)+(-1)^n\left(\begin{array}{c}2 n \\ n\end{array}\right)\right]\\ \\ \displaystyle \qquad =\frac{1}{2^{2 n-1}(-1)^{n}} \sum_{k=0}^{n-1}(-1)^{k}\left(\begin{array}{c}2 n \\ k\end{array}\right) \cos (2 n-2 k) x+\frac{1}{2^{2n}} \left(\begin{array}{c}2 n \\ n\end{array}\right)$$

1. $$\displaystyle \sin ^{2 n+1} x=\left(\frac{z-\bar{z}}{2 i}\right)^{2 n+1}$$

$$\displaystyle \quad \qquad\qquad =\frac{1}{2^{2 n+1}(-1)^{n} i} \sum_{k=0}^{2 n+1}\left(\begin{array}{c}2 n+1 \\ k\end{array}\right) z^{2 n+1-k}(-\bar{z})^{k}$$

$$\displaystyle \quad \qquad\qquad =\frac{1}{2^{2 n+1}(-1)^{n} i} \sum_{k=0}^{n}(-1)^{k}\left(\begin{array}{c}2 n+1 \\ k\end{array}\right)\left(z^{2 n+1-2 k}-\bar{z}^{2 n+1-2 k}\right)$$

$$\displaystyle \quad \qquad\qquad =\frac{1}{2^{2 n+1}(-1)^{n} i} \sum_{k=0}^{n}(-1)^{k}\left(\begin{array}{c}2 n+1 \\ k\end{array}\right) 2 i \sin (2 n+1-2 k) x$$

$$\displaystyle \quad \qquad\qquad =\frac{1}{2^{2 n} (-1)^{n}} \sum_{k=0}^{n}(-1)^{k}\left(\begin{array}{c}2 n+1 \\ k\end{array}\right) \sin (2 n+1-2 k) x$$

• There should be a partial sum formula. (+1) Commented Oct 9, 2021 at 21:00

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.