Integrability of $1/(1+|x|)^{p(x)}$ Let $p(x)$ be a measurable function on $\mathbb{R}$ such that $p(x)>1$. Is it true that the function $1/(1+|x|)^{p(x)}$ is integrable on  $\mathbb{R}$?
 A: $${\bf \text{Hint}}\text{: consider } p(x) = 1 + e^{-|x|}.$$
A: It is not true. To show it, the first thought would be to pick a function $p(x)$ which is $>1$ but decays to $1$ very fast at infinity. As parsiad suggests, $p(x)=1+e^{-|x|}$ works nicely, but then we need to prove that $1/(1+|x|)^{1+e^{-|x|}}$ is not integrable, which didn't look friendly enough for me.
The choice $p(x)=1+1/\log(1+|x|)$
also works ---showing that we actually didn't need such a fast decay to $1$---, and can be quickly verified:
\begin{align}
\int_\mathbb{R}\frac{dx}{(1+|x|)^{1+\frac{1}{\log(1+|x|)}}}&=2\int_0^{+\infty}\frac{dx}{(1+x)^{1+\frac{1}{\log(1+x)}}}\\&=\frac{2}{e}\int_0^{+\infty}\frac{dx}{1+x}=\frac{2}{e}\lim_{x\to+\infty}\log(1+x)=+\infty,
\end{align}
where I've used that $$(1+x)^{1+\frac{1}{\log(1+x)}}=e^{\left(1+\frac{1}{\log(1+x)}\right)\log(1+x)}=e\cdot e^{\log(1+x)}=e(1+x).$$
A: Let $p(x) = 1+1/(|x|+1).$ Then
$$\tag 1 \int_{\mathbb R} \left (\frac{1}{1+|x|}\right)^{p(x)}\, dx  \ge \int_1^\infty \left (\frac{1}{1+x}\right)^{1+1/x}\, dx.$$
Now verify that $(1+x)^{1+1/x} \sim (1+x).$ It follows that the integral on the right of $(1)$ diverges, hence so does the left side of $(1).$.
A: It's obvious that there exists $p(x)>1$ such that $$\frac1{(1+|x|)^{p(x)}}=\frac12\frac1{1+|x|}\quad\quad(x\ne0).$$
