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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 $\mathbb{R}$?

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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 $(1).$.

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$${\bf \text{Hint}}\text{: consider } p(x) = 1 + e^{-|x|}.$$

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    Your example works. But I cannot accept an hint as an answer. – 2017-01-23
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    @Josh: You can accept anything you want as an answer. There are no restrictions. – 2017-01-23
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    I give you an hint: I don't like an hint as an answer. You are not writing a book for students. You are on stack exchange. When I was a student they asked me to solve a problem during an examination. I answered: hint: consider... They didn't appreciate. – 2017-01-23
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    @Josh it's a generally followed idea that when the OP doesn't show much effort into solving their question, to give them assistence but force them to think about the problem a bit. – 2017-01-23
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    @Stella I know what you mean. But this idea is good for students not for stack exchange. A hint can be given as a comment. My aim was to create a small discussion and new observations about that question and it is what I got from the other answers. That an exponential decay was sufficient was a trivial guess. Now I now that a so fast decay is not necessary and I have learned something. – 2017-01-23
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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 $$(1+x)^{1+\frac{1}{\log(1+x)}}=e^{\left(1+\frac{1}{\log(1+x)}\right)\log(1+x)}=e\cdot e^{\log(1+x)}=e(1+x).$$

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    Your exponent function is not greater than 1. You must consider something which is asymptotic to 1 around infinity. – 2017-01-23
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    Ooops how silly of mine. I will delete it and see if I can fix it. Thank you for noticing it!@Josh – 2017-01-23
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    @Josh I think it is ok now. Actually this choice of $p(x)$ was already my first guess. I just thought It was simpler without the '1+' when I was doing the calculations and forgot that it was fundamental. – 2017-01-23