Using contour integration, or other means, is there a way to find a general form for $\int_{0}^{\infty}\frac{\sin^{n}(x)}{x^{n}} \, dx$?

While studying some CA, I ran across methods of evaluating $$\int_0^\infty \frac{\sin x}{x} \, dx, \;\ \int_0^\infty \frac{\sin^2 x}{x^2} \, dx, \;\ \text{and} \ \int_0^\infty \frac{\sin^3 x}{x^{3}} \, dx.$$

Is there a way to find a closed form for $$\int_0^\infty \frac{\sin^n x}{x^n} \, dx, \ n \in \mathbb{N}_{>0} ?$$

Rather it be contour integration or some clever method using real analysis.

• Two comments: (1) "evaluating" or "finding" is a suitable word in this context; "solving" is not. One solves problems; one solves equations; one evalutates expressions. (2) If you write 3 sin x in $\TeX$, it looks like this: $3 sin x$; but if you write 3\sin x, it looks like this: $3\sin x$. The $\sin$ is not italicized but the $x$ is, and proper spacing before and after $\sin$ is there. The latter notation is standard. – Michael Hardy May 18 '12 at 16:37
• Sorry, michael. I know the difference. I usually include the \ in front of the sin as well. Must have been half out of it this morning. – Cody May 18 '12 at 19:35

$$\int_0^\infty \frac{\sin^n(x)}{x^n} \mathrm{d} x = \frac{\pi}{2^{n+1} \cdot (n-1)!} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} (2k-n)^{n-1} \operatorname{sign}(2k-n)$$

where $\operatorname{sign}(x) = \cases{ 1 &$x > 0$\\ 0 &$x = 0$\\ -1 &$x < 0$}$.

As to the (probabilistic) proof, notice that $\frac{\sin(t)}{t}$ is the characteristic function of a uniform random variable on $(-1,1)$. The sum of $n$ independent identically distributed such uniform random variables is known as Irwin-Hall random variable $Y_n$, and the integral in question is a multiple of its PDF evaluated at $x=0$: $$\phi_{Y_n}(x) = \frac{1}{2 \pi} \int_{-\infty}^\infty \frac{\sin^n(t)}{t^n} \mathrm{e}^{-i t x} \mathrm{d} t = \frac{1}{\pi} \int_{0}^\infty \frac{\sin^n(t)}{t^n} \cos(t x) \mathrm{d} t$$ The closed form for the PDF is given on the wikipedia with the reference.

As to more explicit derivation. We first integrate by parts, $n-1$ times, then use binomial theorem for $\sin^n(x)$: $$\begin{eqnarray} \int_0^\infty \frac{\sin^n(x)}{x^n} \mathrm{d} x &=& \int_0^\infty \frac{\mathrm{d}^{n-1}}{\mathrm{d} x^{n-1}}\left( \sin^n(x) \right) \frac{1}{(n-1)!}\frac{\mathrm{d} x}{x} \\ &=& \frac{1}{(n-1)!} \int_0^\infty \frac{1}{2^n i^n} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} \frac{\mathrm{d}^{n-1}}{\mathrm{d} x^{n-1}}\left( \mathrm{e}^{i (2k-n)x} \right) \frac{\mathrm{d} x}{x} \\ &=& \frac{1}{(n-1)!} \int_0^\infty \frac{1}{2^n i^n} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} \left(i (2k-n)\right)^{n-1} \mathrm{e}^{i (2k-n)x} \frac{\mathrm{d} x}{x} \\ &=& \frac{1}{(n-1)!} \int_0^\infty \frac{1}{2^n} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} \left((2k-n)\right)^{n-1} \sin((2k-n)x) \frac{\mathrm{d} x}{x} \end{eqnarray}$$ In the last line, $\mathrm{e}^{i (2k-n) x}$ was expanded use Euler's formula, and since the sum is real, only real summands are retained. Then, integrating term-wise nails it: $$\begin{eqnarray} \int_0^\infty \frac{\sin^n(x)}{x^n} \mathrm{d} x &=& \frac{1}{2^n} \frac{1}{(n-1)!} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} \left((2k-n)\right)^{n-1} \int_0^\infty \sin((2k-n)x) \frac{\mathrm{d} x}{x} \\ &=& \frac{1}{2^n} \frac{1}{(n-1)!} \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} \left((2k-n)\right)^{n-1} \frac{\pi}{2} \operatorname{sign}(2k-n) \end{eqnarray}$$

• Wow, thanks Sasha. Great insight. Thanks very much. – Cody May 18 '12 at 19:37
• BTW Sasha, thanks to your closed form I know what I am shooting for. I am going to try to arrive at it using contour integration. I doubt if I am successful, but I will give it an earnest effort. – Cody May 18 '12 at 21:28

I have a generalized elementary method for this problem，If f (x) is an even function, and the period is $\pi$,we have: $$\int_{0}^\infty f(x)\frac{\sin^nx}{x^n}dx=\int_{0}^\frac{\pi}{2}f(x)g_n(x)\sin^nxdx \qquad (1)$$

Where the $g_n(x)$ in (1) is as follows $$g_n(x)=\begin{cases}\frac{(-1)^{n-1}}{(n-1)!}\frac{d^{n-1}}{dx^{n-1}}\left(\csc x\right),& \text{for n is odd n\in\Bbb N and}\\[2ex] \frac{(-1)^{n-1}}{(n-1)!}\frac{d^{n-1}}{dx^{n-1}}\left(\cot x\right),& \text{ for n is even .} \end{cases}$$ —————————————————————————————————————————————————— Proof: \begin{align} \int_{0}^\infty f(x)\frac{\sin^nx}{x^n}dx&=\sum_{k=0}^\infty\int_{k\pi}^{(2k+1)\frac{\pi}{2}}f(x)\left(\frac{\sin x}{x}\right)^ndx+\sum_{k=1}^\infty\int_{(2k-1)\frac{\pi}{2}}^{k\pi}f(x)\left(\frac{\sin x}{x}\right)^ndx\\ &=\sum_{k=0}^\infty\int_{0}^{\frac{\pi}{2}}f(x+k\pi)\left(\frac{\sin (x+k\pi)}{x+k\pi}\right)^ndx+\sum_{k=1}^\infty\int_{-\frac{\pi}{2}}^{0}f(x+k\pi)\left(\frac{\sin (x+k\pi)}{x+k\pi}\right)^ndx\\ &=\sum_{k=0}^\infty(-1)^{nk}\int_{0}^{\frac{\pi}{2}}f(x)\left(\frac{\sin x}{x+k\pi}\right)^ndx+\sum_{k=1}^\infty(-1)^{nk}\int_{0}^{\frac{\pi}{2}}f(-x)\left(\frac{\sin x}{x-k\pi}\right)^ndx\\ &=\int_{0}^{\frac{\pi}{2}}f(x)\sin^nx\left(\frac{1}{x^n}+\sum_{k=1}^\infty(-1)^{nk}\left[\frac{1}{(x+k\pi)^n}+\frac{1}{(x-k\pi)^n}\right]\right)dx\\ &=\int_{0}^{\frac{\pi}{2}}f(x)\sin^nxg_n(x)dx \end{align} 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 n-1 order derivative,thus we obtain $g_n(x)$. —————————————————————————————————————————————————— 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}\\ (4.) \int_{0}^{\infty}\cos 2xy\frac{\sin^2x}{x^2}dx &=\int_{0}^{\frac{\pi}{2}}\cos 2xydx =\frac{\sin\pi y}{2y}\\ \end{align} \begin{align} \end{align}

• I'm poor at writing formulas. Can someone help me fix them? – JamesJ Aug 25 '17 at 13:52
• @Cody,Why didn't anyone pay attention to the answer? – JamesJ Aug 25 '17 at 16:43