I'm confused as to why $$\int_{-\infty}^\infty\frac{1}{x}dx$$ diverges. If $\frac{1}{x}$ is an odd function shouldn't the area to the left of origin be the opposite of the area to the right of origin, resulting in a net area of 0?

I am wondering whether there exists a function such that: $$\lim_{x \rightarrow a}f(x)=\infty$$ at some point $a$ on the real axis but yet, $$\int_{-\infty}^{+\infty}\left|f(x)\right|\ dx<\infty$$ Does the fact that a function is unbounded imply that it has no finite integral?

Is the following theorem true for multivariate case? Convergence in distribution plus continuity of limiting distribution implies uniform convergence. Remark: 1) In univariate case, one can find A such that F(x) vanishes outside [-A, A], and make use of uniform continuity of F on [-A, A](cont...

Define $\|.\|_{T^k(\Omega)}$ as $$ \|f\|_{T^k(\Omega)} = \|f\|_{L^2(\Omega)} + \|(\sum\limits_{i=1}^d(\frac{\partial^{k}f}{\partial x_i^{k}})^2)^{\frac{1}{2}}\|_{L^2(\Omega)} $$ Is this norm equivalent to Sobolev norm, $\|.\|_{W^{k,2}(\Omega)}$, under two cases $\Omega$ is a bounded open ...