Let $a,p$ be real numbers and $a\gt1$.Which of the following statements are true?
$1$.If $p\gt 1,\int_{-\infty}^{\infty} \frac{1}{|x|^{pa}} dx \lt \infty $
$2$.If $p\gt \frac{1}{a}$ then $\int_{-\infty}^{\infty} \frac{1}{|x|^{pa}} dx \lt \infty $
$3$.If $p\lt \frac{1}{a}$ then $\int_{-\infty}^{\infty} \frac{1}{|x|^{pa}} dx \lt \infty $
$4$.For any $p\in \Bbb R$ we have $\int_{-\infty}^{\infty} \frac{1}{|x|^{pa}} dx = \infty $
For this question if I separate the limit of the integral from $-\infty$ to $0$ and $0$ to $\infty$.
$\int_{-\infty}^{\infty}\frac{1}{|x|^{pa}} dx = \int_{-\infty}^{0} \frac{1}{|x|^{pa}} dx $+ $\int_{0}^{\infty} \frac{1}{|x|^{pa}} dx $
$-\int_{-\infty}^{0} \frac{1}{x^{pa}} dx$ + $\int_{0}^{\infty} \frac{1}{x^{pa}} dx $.After solving this I am facing problem in limit substitution.
For $pa\neq 1$ You have: $$\int_{-\infty}^{\infty}\frac{1}{|x|^{pa}}dx=2\lim_{T\rightarrow\infty}\lim_{t\rightarrow 0}(\frac{1}{1-pa}|T|^{1-pa}-\frac{1}{1-pa}|t|^{1-pa})$$ and You have $$\lim_{t\rightarrow 0}(\frac{1}{1-pa}|T|^{1-pa}-\frac{1}{1-pa}|t|^{1-pa})=\infty$$ for any $T$ if $1-pa<0$ and $$\lim_{t\rightarrow 0}(\frac{1}{1-pa}|T|^{1-pa}-\frac{1}{1-pa}|t|^{1-pa})=\frac{1}{1-pa}|T|^{1-pa}$$ if $1-pa>0$.
If $pa=1$ You have $$\int_{-\infty}^{\infty}\frac{1}{|x|^{pa}}dx=2\lim_{T\rightarrow\infty}\lim_{t\rightarrow 0}(\ln|T|-\ln|t|).$$ Can You take it from here?