Does $\int_{-\infty}^{\infty}\sin^{-1}\left(\frac{\sin(x)}{x}\right)dx$ converge?

Question

My attempts: $$\frac{dy}{dx}=\sin^{-1}\left(\frac{\sin(x)}{x}\right)$$ $$\sin\left(\frac{dy}{dx}\right)=\frac{\sin(x)}{x}$$ $$x\sin\left(\frac{dy}{dx}\right)=\sin(x)$$ Seems like a dead end. $$\int_{-\infty}^{\infty}\sin^{-1}\left(\frac{\sin(x)}{x}\right)dx$$ $$2\int_{0}^{\infty}\sin^{-1}\left(\frac{\sin(x)}{x}\right)dx$$ $$2\int_{1}^{0}\sin^{-1}(u)\frac{du}{\frac{x\cos(x)-\sin(x)}{x^2}}$$ Also seems like a dead end because I can't figure out how to solve for $x$ in $u=\frac{\sin x}{x}$.

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Edit: I also see that $\left|\frac{\sin(x)}{x}\right|\le\left|\sin^{-1}\left(\frac{\sin(x)}{x}\right)\right|\le\left|\frac{\pi}{2}\frac{\sin(x)}{x}\right|$ but I doubt anything can be done with that.

Answer 1

1. Using the Taylor expansion of $\arcsin x$, it is easy to check that there exists a continuous function $g : [-1, 1] \to \mathbb{R}$ satisfying $$ \arcsin x = x + x^3 g(x), \qquad \forall x \in [-1, 1]. \tag{1} $$

2. Let $ \operatorname{sinc}(x) = \frac{\sin x}{x}$ and $\operatorname{sinc}(0) = 1$. Then it is easy to check that $\left|\operatorname{sinc}(x)\right| \leq 1$ for all $x \in \mathbb{R}$. Moreover, the improper integral of $\operatorname{sinc} x$ on $(-\infty, \infty)$ converges, that is, $$ \int_{-\infty}^{\infty} \operatorname{sinc}(x) \, \mathrm{d}x = \lim_{\substack{a \to -\infty \\ b \to \infty}} \int_{a}^{b} \operatorname{sinc}(x) \, \mathrm{d}x $$ converges.

3. From $\text{(1)}$, it follows that $\left|\arcsin(\operatorname{sinc}x) - \operatorname{sinc}x \right| \leq \frac{C}{(1+|x|)^3}$ for some absolute constant $C > 0$. This tells that $$ \int_{-\infty}^{\infty} \left[ \arcsin(\operatorname{sinc} x) - \operatorname{sinc} x \right] \, \mathrm{d}x $$ converges absolutely by the comparison test.

Combining these facts, it follows that $\int_{-\infty}^{\infty} \arcsin(\operatorname{sinc} x) \, \mathrm{d}x$ converges.

Answer 2

First of all we check the behaviour of the integrand for small $x$. The ratio $\sin(x)/x$ is then slightly smaller than one, so we can indeed take the arcsin and obtain the limit value of $\pi /2$. Hence there is no problem for $x$ around zero.

Next we check the behaviour for large values of $x$. It is convenient to rewrite the formula for the integrand $f = \arcsin(\sin(x)/x))$ as $\sin(f) = \sin(x)/x$. We consider an interval of width $2\pi$ around the central value $x_0 = (2N + 1/2)\pi$, with $N$ large. Then the RHS is given by:

$$RHS = \sin((2N+1/2)\pi +\delta)/((2N+1/2)\pi+\delta) = \cos(\delta)/(x_0+\delta)$$

where $\delta$ is defined as $\delta = x-x_0$. Since $\delta$ is small compared to $x_0$ we can expand the denominator up to third order in $x_0$:

$$RHS = \cos(\delta)(1/x_0 - \delta/x_0^2 + \delta^2/x_{0}^{3})$$

The LHS can be expanded for small values of $f$ as $LHS = \sin(f) = f - f^3 /6$. Combining these results we obtain, correct up to third order in $1/x_0$:

$$f = \cos(\delta)(1/x_0 - \delta/x_0^2 + \delta^2/x_{0}^{3}) + \cos(\delta)^3/6x_0^3$$

We now integrate over the interval $(-\pi, \pi)$. We see that the first order term vanishes (integrating a cosine over $2\pi$) and also the second order term (antisymmetric function). Hence the integral yields a contribution of third order and higher in $x_0$. Contributions from successive intervals can therefore be summed without difficulty and the result converges. Therefore the integral as a whole converges.