For which $p>1$ and $d>1,$ we can expect $\int_0^{\infty} \frac{r^{d-1}}{ (1+r^2)^{(d+1)p/2}} dr< \infty$

Question

Let $1\leq p \leq \infty, d \in \mathbb N$

My question is: For which $p>1$ and $d>1,$ we can expect $\int_0^{\infty} \frac{r^{d-1}}{ (1+r^2)^{(d+1)p/2}} dr< \infty$

I think: For $p=d=1, $ we have $\int_0^{\infty} \frac{1}{1+r^2} dr= \lim_{R\to \infty}\int_0^R \frac{1}{1+r^2} dr = \frac{\pi}{2}.$

Answer 1

It converges for all $p,d\geq 1$:

Since the denominator tends to $1$ as $ r \to 0$ the integral from $0$ to $1$ is finite. Now look at the behaviour on $(1,\infty)$. What you need is convergence of $\int_1^{\infty} \frac 1 {r^{(d+1)p+1-d}}dr$ which is true because $(d+1)p+1-d >1$.