I am working on a problem which have lead me to the following integral:
$$\lim_{\epsilon \rightarrow 0^{+}} \epsilon^{8}\int_{E_{\epsilon}}\parallel{x} \parallel ^{-12}\cos(x_{1})dx \, , $$ where$E_{\epsilon } =\left\{ x=(x_{1},x_{2},x_{3},x_{4})\in R^{4} \, : \, \parallel x\parallel\geq \epsilon \, \right\}$.The norm should be regarded as euclidean norm. I am trying to check the exisitence of the limit using integration by part, but still can't give a clear answer. If it converges, please help give the value.
$$\int_{ E_\epsilon}\|x\|^{-12} \cos ( x_1 ) dx = \int_{ E_\epsilon}\|x\|^{-12}(1-(1- \cos ( x_1 )) dx=\int_{ E_\epsilon}\|x\|^{-12} dx-\int_{ E_\epsilon}\|x\|^{-12}(1- \cos ( x_1 )) dx$$ We know from @Daniel Fischer that $\lvert 1- \cos (x_1)\rvert \leqslant \frac{1}{2} x_1^2 \leqslant \frac{1}{2}\lVert x\rVert^2$. Therefore,$$ |\int_{ E_\epsilon}\|x\|^{-12}(1- \cos ( x_1 )) dx|\leq \frac{1}{2}\int_{ E_\epsilon}\|x\|^{-10} dx$$ Hence,$$\int_{ E_\epsilon}\|x\|^{-12} dx-\frac{1}{2}\int_{ E_\epsilon}\|x\|^{-10} dx\leq\int_{ E_\epsilon}\|x\|^{-12} \cos ( x_1 ) dx\leq\int_{ E_\epsilon}\|x\|^{-12} dx+\frac{1}{2}\int_{ E_\epsilon}\|x\|^{-10} dx$$ $$lim_{\epsilon \rightarrow 0^{+}}\epsilon^{8} (\frac{\pi^2 }{4\epsilon ^8} -\frac{\pi^2 }{3\epsilon ^6})\leq lim_{\epsilon \rightarrow 0^{+}} \epsilon^{8}\int_{E_{\epsilon}}\parallel{x} \parallel ^{-12}\cos(x_{1})dx\leq lim_{\epsilon \rightarrow 0^{+}}\epsilon^{8} (\frac{\pi^2 }{4\epsilon ^8} +\frac{\pi^2 }{3\epsilon ^6})$$ $$\int_{ E_\epsilon}\|x\|^{-12} \cos ( x_1 ) dx=\frac{\pi^2 }{4}$$ Thanks for all your help.