Is $p \mapsto \int_0^\infty \frac{1}{1 + x^p} dx$ continuous? Let $\varphi: \mathbb{R}^+ \cup \{0, \infty\} \to \mathbb{R} \cup \{\infty\}$, $\varphi(p) = \int_0^\infty \frac{1}{1 + x^p} dx$. Is $\varphi$ a continuous function?
It seems so, since as $p$ gets larger, $\varphi(p)$ gets smaller continuously. Moreover, $\varphi(p) = \infty$ for $0 \leq p \leq 1$, but $\varphi(0) = \int_0^\infty 1 dx$ and $\varphi(1) = \int_0^\infty \frac{1}{1+x} dx = [ln (1+x)]_0^\infty$, so $\varphi(1)$ diverges to $\infty$ "slower" than $\varphi(0)$.
I tried to use the epsilon-delta argument, but I don't think it works because the integral from $0$ to $\infty$ keeps giving me $\epsilon \cdot \infty$.
 A: First fix $R>0$ and look at $\varphi_R(p):=\int_0^R \frac{1}{1+x^p}dx$. This is continuous since $f(x,p)=\frac{1}{1+x^p}$ is and primitives of continuous functions are continuous. If we can show that $\varphi_N\overset{N\rightarrow \infty}{\longrightarrow} \varphi$ uniformly we are finished, since the uniform limit of a sequence of continuous functions itself is continuous.
Thus we have to show that
\begin{align*}  \sup_{p\in I} |\varphi(p)-\varphi_N(0)|:=\sup_{p\in I} \left|\int_0^\infty \frac{1}{1+x^p}dx-\int_0^N \frac{1}{1+x^p}dx\right|=\sup_{p\in I} \left|\int_N^\infty \frac{1}{1+x^p}dx\right|\end{align*}
goes to zero, as $N\rightarrow \infty$. Here it becomes important what interval $I$ we choose. As you have already noticed, $\varphi(p)=\infty$ if $0\leq p\leq 1$. Hence fix $\varepsilon>0$ and denote $I:=[1+\varepsilon, \infty)$, which gives us some comforting distance from $p=1$.
For that interval $I$ we can proceed with the following estimation:
\begin{align*}  \sup_{p\in I} \left|\int_N^\infty \frac{1}{1+x^p}dx\right|\leq \sup_{p\in I}\int_N^\infty \frac{1}{x^p}dx=\sup_{p\in I}\frac{1}{1-p}N^{1-p}\leq\sup_{p\in I}\frac{1}{\varepsilon}N^\frac{1}{\varepsilon}=\frac{1}{\varepsilon}N^\frac{1}{\varepsilon}\longrightarrow 0.\end{align*}
Hence for each $\varepsilon>0$ this convergence is uniform and $\varphi$ is continuous on every interval $[1+\varepsilon,\infty)$ or equivalently, $\varphi$ is continuous on $(1,\infty)$.
A: Yes, it is continuous. First, check the continuity at $p=1$. We have that
$$\varphi(1)=\int_{0}^{\infty}\frac{dx}{1+x}=+\infty$$
but for ever $p>1$
$$\varphi(p)=\int_{0}^{\infty}\frac{dx}{1+x^p}=\frac{\pi  \csc \left(\frac{\pi }{p}\right)}{p}\longrightarrow +\infty$$
as $p\rightarrow 1^{+}$.
When $p>1$ the integral is continuous because $\phi(p)=\frac{\pi  \csc \left(\frac{\pi }{p}\right)}{p}$ which is continuous for all $p>1$.
Another way to see the continuity for $p>1$ is to observe the fact that the integrand is an $L^{1}([0,\infty[)$ function.  So, for every $q>1$, we have by the dominated convergence theorem that
$$\lim_{p\rightarrow q}\varphi(p)=\lim_{p\rightarrow q}\int_{0}^{\infty}\frac{dx}{1+x^p}=\int_{0}^{\infty}\frac{dx}{1+x^q}=\varphi(q)$$.
