Let $p>0$. For which $p$ is the integral $\int_{\mathbb{R}} \frac{1}{(1+|x|)^p} d\mathcal{L}(x)$ finite?
I assume $f(x)= \frac{1}{(1+|x|)^p}$ is integrable for $p>1$, but how exactly can I show this?
Let $p>0$. For which $p$ is the integral $\int_{\mathbb{R}} \frac{1}{(1+|x|)^p} d\mathcal{L}(x)$ finite?
I assume $f(x)= \frac{1}{(1+|x|)^p}$ is integrable for $p>1$, but how exactly can I show this?
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You can do the following: split the integral al $2$ then
\begin{align*}\int_{\mathbb{R}}\frac{1}{(1+|x|)^p}\,d\mathcal{L}&=2\left( \int_{0}^2\frac{1}{(1+x)^p}\,d\mathcal{L}+ \int_{2}^\infty\frac{1}{(1+x)^p}\,d\mathcal{L}\right)\\ &=2\left( \int_{0}^2\frac{1}{(1+x)^p}\,d\mathcal{L}+ \int_{\color{red}3}^\infty\frac{1}{x^p}\,d\mathcal{L}\right).\end{align*} The first integral is finite and smaller than $2$ and the convergence of the second depends on $p$ as you said, but the antiderivative is now easy to compute.
Previous reasoning:
For $|x|\leq c$ (c any constant) the integrand is finite so the intergral cannot diverge.
For $|x|\gg 1$ $\frac{1}{(1+|x|)^p}\approx \frac{1}{|x|^p}$ which is easy to integrate.