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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
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2 Answers

As commented by sos440 the integral doesn't converge because the integrand doesn't tend to $0$ as $x\rightarrow \pm\infty$

$$\begin{eqnarray*} \lim_{x\rightarrow +\infty }\frac{1}{x\sin \frac{1}{x}} &=&\lim_{x \rightarrow +\infty }\frac{1/x}{\sin \frac{1}{x}}=\lim_{y\rightarrow 0^{+}} \frac{y}{\sin y}=1, \\ \lim_{x\rightarrow -\infty }\frac{1}{x\sin \frac{1}{x}} &=&\lim_{x \rightarrow -\infty }\frac{1/x}{\sin \frac{1}{x}}=\lim_{y\rightarrow 0^{-}} \frac{y}{\sin y}=1. \end{eqnarray*}$$

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I have the following differential equation dy/dt=f( y ),so I had to solve it and I got this integral,where f(x) is the above mentioned function –  Maria Aug 26 '12 at 20:05
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
@Robert Israel: Thanks! I've corrected. –  Américo Tavares Aug 26 '12 at 20:09
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
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This integral can be solved by a trick. The point at x=0 is a essential singularity, so the convention for f(0)=0 make sense by interpretate this integral as a Lebesgue Integral: Using the substitution $x=1/u$, and take the limit $r \to \infty$

$\int_{-\infty}^{+\infty} {{dx} \over {x \sin (1/x)}} = \lim_{r \to \infty} \int_{-r}^{+r} {{dx} \over {x \sin(1/x)}} = \int_{-1/r}^{1/r} { {- u} \over {\sin(u) u^2}} du = $ $ \int_{-1/r}^{1/r} { \csc(u) \over u} du + \pi i \times Res (\csc(u)/u , u=0) $

The residue term is justified by change of variables, and the new integration path around the singularity at u=0. However, the new function don't have a essential singularity anymore, and the calculation of the residue is simple: equals 0. (The function is even, so it don't have non-zero odd terms in Laurent series).

Taking the limit, the integral equals to zero:

$\int_{-\infty}^{+\infty} {{dx} \over {x \sin (1/x) }} = \lim_{r \to \infty} \int_{-1/r}^{1/r} {\csc(u) \over u} du = \int_0^0 {\csc(u) \over u} du = 0 $

This means in the context of problem, the integral should be zero, when we apply the Lebesgue Integral and taking the Principal Value, but the integral itself are divergent since it fail any convergence tests.

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