# Nonelementary integration $\int_{-\infty}^\infty 1/(x\sin(1/x)) dx$

Given the function

$$f(x)= \cases{ x\sin\Big(\frac{1}{x}\Big) & if x\neq 0 \\ 0 & if x=0}$$

Find $$\int\limits_{-\infty}^\infty \frac{1}{f(x)} dx$$

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I partly fixed up the TeX, but the question does not make full sense, since $f(x)$ is not mentioned in the integral. Can you (by editing or leaving a comment) say what the question really is? – André Nicolas Aug 26 '12 at 19:28
Is there any particular reason to believe that it converges? The integrand is $$1+O\left(\frac{1}{x^2}\right)$$ as $x\to\pm\infty$, which already shows that the integral does not converge, much less its highly singular behavior near the origin. – sos440 Aug 26 '12 at 19:55
@sos440 it doesn't converge. You should write an answer. – user2468 Aug 26 '12 at 19:57
@JenniferDylan, I suspect that the questioner might have made a mistake. So I'm going to wait for a while to see if it is true. It won't be too late then to post an answer. – sos440 Aug 26 '12 at 20:04
Someone nearly simultaneously posted the indefinite integral version of the problem on mathoverflow: mathoverflow.net/questions/105555/integrating-1-xsin1-x-closed – Ben Crowell Aug 26 '12 at 20:04

As commented by sos440 the integral doesn't converge because the integrand doesn't tend to $0$ as $x\rightarrow \pm\infty$
$$\begin{eqnarray*} \lim_{x\rightarrow +\infty }\frac{1}{x\sin \frac{1}{x}} &=&\lim_{x \rightarrow +\infty }\frac{1/x}{\sin \frac{1}{x}}=\lim_{y\rightarrow 0^{+}} \frac{y}{\sin y}=1, \\ \lim_{x\rightarrow -\infty }\frac{1}{x\sin \frac{1}{x}} &=&\lim_{x \rightarrow -\infty }\frac{1/x}{\sin \frac{1}{x}}=\lim_{y\rightarrow 0^{-}} \frac{y}{\sin y}=1. \end{eqnarray*}$$
In the second one, be careful: $\sin(1/x)$ is not always positive. It would be true on, say, $[1,\infty)$ though. – Robert Israel Aug 26 '12 at 20:06
@Maria: for the differential equation you don't need the integral from $-\infty$ to $\infty$. – Robert Israel Aug 26 '12 at 20:08
The differential equation $dy/dt = f(y)$ has constant solutions $y = k$ whenever $f(k)=0$. In between those you have solutions $y(t)$ defined implicitly by $t - t_0 = \int_{y_0}^y dy/f(y)$ (an integral that can't be done in closed form). If $k_1 < y_0 < k_2$ where $f(k_1) = f(k_2) = 0$ but $f(y) \ne 0$ for $y$ between $k_1$ and $k_2$, this solution will have $y$ either increasing or decreasing (depending on the sign of $f(y)$ in this interval) and approaching one of $k_1$ and $k_2$ as $t \to -\infty$ and the other as $t \to +\infty$. – Robert Israel Aug 26 '12 at 20:18