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
$$\int\limits_{-\infty}^\infty \frac{1}{f(x)} dx $$
 A: As commented by sos440 the integral doesn't converge because the integrand doesn't tend to $0$ as $x\rightarrow \pm\infty$  
$$\begin{eqnarray*}
\lim_{x\rightarrow +\infty }\frac{1}{x\sin \frac{1}{x}} &=&\lim_{x
\rightarrow +\infty }\frac{1/x}{\sin \frac{1}{x}}=\lim_{y\rightarrow 0^{+}}
\frac{y}{\sin y}=1, \\
\lim_{x\rightarrow -\infty }\frac{1}{x\sin \frac{1}{x}} &=&\lim_{x
\rightarrow -\infty }\frac{1/x}{\sin \frac{1}{x}}=\lim_{y\rightarrow 0^{-}}
\frac{y}{\sin y}=1.
\end{eqnarray*}$$
A: 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.
