Evaluate $\int_{-\infty}^\infty \frac{\sin{(x\sin{x})}}{x^2}dx$.

Question

Evaluate $$\int_{-\infty}^\infty \frac{\sin{(x\sin{x})}}{x^2}dx.$$

Numerical investigation suggests that it equals $\pi$.

Let $f(\alpha)=\frac{1}{\pi}\int_{-\alpha}^\alpha \frac{\sin{(x\sin{x})}}{x^2}dx$.

Wolfram (free version) gives me the following results:

$f(10)=0.992293$
$f(20)=0.996916$
$f(30)=0.998604$
$f(100)=0.999761$
$f(200)=0.999916$
$f(300)=0.999934$

But with even larger values of $\alpha$, Wolfram behaves erratically:

$f(1\times10^5)=2.12907$
$f(2\times10^5)=0.574468$
$f(3\times10^5)=0.351411$

And finally, Wolfram gives:

$f(\infty)=0.998775$

But in the comments, @NN2 said that the paid version of Wolfram gives some detailed warnings about the accuracy of this last result.

Perhaps someone with more reliable computing power could shed some light on $f(\infty)$.

My attempt

I tried to use some of the techniques for proving $\int_0^\infty {\sin{x}\over x}dx=\pi/2$, to no avail. For example, this one sets up $I(a)=\int_0^\infty e^{-ax}\frac{\sin x}{x}dx$, then evaluates $I'(a)$ via integration by parts. If we just have $\sin{x}$ this is easy, but the fact that $\int_0^\infty{\sin{(x\sin{x})}\over x^2}dx$ involves a sine within a sine, seems to make everything difficult.

Context

I've been searching (in vain) for a definite integral with no numbers that equals $\pi/7$, and I stumbled upon this integral. I was surprised that it seems to admit a closed form, $\pi$.

Answer 1

Using the known expansion for $\sin(x\sin x)$ in terms of Bessel functions, $$ \int_{ - \infty }^{ + \infty } {\frac{{\sin (x\sin x)}}{{x^2 }}{\rm d}x} = 2\int_0^{ + \infty } {\frac{{\sin (x\sin x)}}{{x^2 }}{\rm d}x} \\ = 4\sum\limits_{k = 0}^\infty {( - 1)^k \int_0^{ + \infty } {\frac{{J_{2k + 1} (x)\sin ((2k + 1)x)}}{{x^2 }}{\rm d}x} } . $$ The term corresponding to $k=0$ is, by $(10.16.1)$ and $(10.22.57)$, $$ \int_0^{ + \infty } {\frac{{J_1 (x)\sin (x)}}{{x^2 }}{\rm d}x} = \sqrt {\frac{\pi }{2}} \int_0^{ + \infty } {\frac{{J_1 (x)J_{1/2} (x)}}{{x^{3/2} }}{\rm d}x} = \sqrt {\frac{\pi }{2}} \frac{1}{{\sqrt 2 }}\frac{{\Gamma \!\left( {\frac{1}{2}} \right)\Gamma\! \left( {\frac{3}{2}} \right)}}{{2\Gamma (1)\Gamma\! \left( {\frac{3}{2}} \right)}} = \frac{\pi }{4}. $$ For $k\ge 1$, $(10.22.56)$ gives \begin{align*} & \int_0^{ + \infty } {\frac{{J_{2k + 1} (x)\sin ((2k + 1)x)}}{{x^2 }}{\rm d}x} = \sqrt {\frac{{\pi (2k + 1)}}{2}} \int_0^{ + \infty } {\frac{{J_{2k + 1} (x)J_{1/2} ((2k + 1)x)}}{{x^{3/2} }}{\rm d}x} \\ & = \frac{{\sqrt{\pi}\,\Gamma \!\left( {k + \frac{1}{2}} \right)}}{{4 (2k + 1)^{2k} \Gamma (1 - k)}}{\bf F}\!\left( {k + \frac{1}{2},k;2k + 2;\frac{1}{{(2k + 1)^2 }}} \right) = 0, \end{align*} since $1/\Gamma(z)$ is zero for $z=0,-1,-2,\ldots$. Here $\bf{F}$ denotes the regularised hypergeometric function. Hence the value of the original integral is indeed $\pi$.

Answer 2

Let $$S_n=2\int_{n \pi}^{(n+1)\pi} \frac{\sin (x \sin (x))}{x^2}\,dx \qquad \text{and} \qquad T_p=\sum_{n=0}^p S_n$$ and compute for an high accuracy (the numbers below are truncated)

$$\left( \begin{array}{cc} p & T_p \\ 0 & 3.08335068793 \\ 1 & 3.08795887316 \\ 2 & 3.11823682853 \\ 3 & 3.12373609535 \\ 4 & 3.13000260857 \\ 5 & 3.13207351895 \\ 6 & 3.13445795667 \\ 7 & 3.13547222096 \\ 8 & 3.13665697958 \\ 9 & 3.13723885677 \\ 10 & 3.13792351502 \\ 20 & 3.14019199333 \\ 30 & 3.14081055776 \\ 40 & 3.14107819187 \\ 50 & 3.14122173949 \\ 60 & 3.14130906854 \\ 70 & 3.14136680535 \\ 80 & 3.14140730433 \\ 90 & 3.14143699830 \\ 100 & 3.14145953226 \\ 200 & 3.14154523773 \\ 300 & 3.14156678031 \\ 400 & 3.14157582788 \\ 500 & 3.14158060530 \\ 600 & 3.14158348369 \\ 700 & 3.14158537420 \\ \end{array} \right)$$

I did not see a single error or warning message from Mathematica

I stop at this point this @Gary provided the proof.

Answer 3

With regard to Gary's answer, we can also use contour integration to show that $$\int_{0}^{\infty} \frac{J_{1}(x) \sin(x)}{x^{2}} \, \mathrm dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{J_{1}(x) \sin(x)}{x^{2}} \, \mathrm dx= \frac{\pi}{4}$$ and $$\int_{0}^{\infty} \frac{J_{2k+1}(x) \sin\left((2k+1)x\right)}{x^{2}} \, \mathrm dx = \frac{1}{2} \int_{-\infty}^{\infty}\frac{J_{2k+1}(x) \sin\left((2k+1)x\right)}{x^{2}} \, \mathrm dx = 0 $$ for $k \in \mathbb{Z}_{\ge 1}$ by integrating the functions $$f(z,k) = \frac{J_{2k+1}(z) \exp(i(2k+1)z)}{z^{2}}, \, \quad k \in \mathbb{Z}_{\ge 0}, $$ around a contour that consists of the real axis and the semicircle above it.

The functions are analytic in the upper half-plane.

But the contour needs to be indented at the origin for the function $f(z, 0)$ because there is a simple pole there.

Also, the integrals vanish along the semicircle as its radius goes to infinity because the growth in the magnitude of $J_{2k+1}(z)$ as $\Im(z) \to +\infty$ is neutralized by the decay in the magnitude of $\exp(i(2k+1)z)$ as $\Im(z) \to +\infty$. (See here.)

Integrating the function $f(z,0)$ around the contour, we get $$\operatorname{PV} \int_{-\infty}^{\infty} \frac{J_{1}(x) \exp\left(ix\right)}{x^{2}} \, \mathrm dx - i \pi \operatorname{Res}[f(z,0),0] =0. $$

Therefore, $$ \begin{align} \operatorname{PV} \int_{-\infty}^{\infty} \frac{J_{1}(x) \exp\left(ix\right)}{x^{2}} \, \mathrm dx &= i \pi \operatorname{Res}[f(z,0),0] \\ &= i \pi \lim_{z \to 0} \frac{J_{1}(z)}{z} \, \exp\left(iz\right) \\ &= i \pi \left(\frac{1}{2} \right)(1) \\& = \frac{i \pi}{2}. \end{align}$$

Equating the imaginary parts on both sides of the equation, we get $$ \int_{-\infty}^{\infty} \frac{J_{1}(x) \sin(x)}{x^{2}} \, \mathrm dx = \frac{\pi}{2}. $$

And integrating the functions $f(z,k)$, $k \in \mathbb{Z}_{\ge 1}$, around the contour, we have simply $$ \int_{-\infty}^{\infty} \frac{J_{2k+1}(x) \exp\left(i(2k+1)x\right)}{x^{2}} \, \mathrm dx =0. $$

Equating the imaginary parts on both sides of the equation, we get $$\int_{-\infty}^{\infty}\frac{J_{2k+1}(x) \sin\left((2k+1)x\right)}{x^{2}} \, \mathrm dx =0 \, , \quad k \in \mathbb{Z}_{\ge 1}.$$