Problem: For $\displaystyle{p \geqslant 1}$, consider \begin{align*} f ( x ) = \frac{1}{ \displaystyle{ x^{ \frac{1}{p} } \big( \log^{2}{x} + 1 \big) } } \qquad \text{for all } x \in ( 0 , \infty). \end{align*} Show that: \begin{align*} f \in L^{p} \big( ( 0 , \infty ) , m \big) \qquad \text{and} \qquad f \notin L^{q} \big( ( 0 , \infty ) , m \big) \text{ for all } q \neq p \text{ and } q \geqslant 1. \end{align*}
Solution for that $f\in L^{p}$: Firstly, we check that: \begin{align*} || f ||_{p}^{p} = \int\limits_{(0, \infty)} | f ( x ) |^{p} d m ( x ) = \int\limits_{(0, \infty)} \left| \frac{1}{x^{ \frac{1}{p} } ( \log^{2}{x} + 1 ) } \right|^{p} d m ( x ) = \underbrace{ \int\limits_{(0, \infty)} \frac{1}{x ( \log^{2}{x} + 1 )^{p} } d m ( x ) }_{\text{Lebesgue-Integral}} = \underbrace{ \int\limits_{0}^{\infty} \frac{1}{x ( \log^{2}{x} + 1 )^{p} } d x }_{\text{Riemann-Integral}} = \\ = \int\limits_{- \infty}^{\infty} \frac{1}{( y^{2} + 1 )^{p}} d y \quad \Big( \text{we set } y = \log{x} \Big) \quad \leqslant \int\limits_{- \infty}^{\infty} \frac{1}{y^{2} + 1} d y = \Big[ \arctan{y} \Big]_{y \to -\infty}^{y \to \infty} = \pi < \infty \Longrightarrow f \in L^{p} \big( ( 0 , \infty ) , m \big). \end{align*}
My attempt to show that $f \notin L^{q}$: Secondly, we have the integral \begin{align*} || f ||_{p}^{q} = \int\limits_{(0, \infty)} | f ( x ) |^{q} d m ( x ) = \int\limits_{(0, \infty)} \left| \frac{1}{x^{ \frac{1}{p} } ( \log^{2}{x} + 1 ) } \right|^{q} d m ( x ) = \int\limits_{(0, \infty)} \left| \frac{1}{x^{ \frac{1}{p} } ( \log^{2}{x} + 1 ) } \right|^{q} d m ( x ) = \\ = \underbrace{ \int\limits_{(0, \infty)} \frac{1}{x^{ \frac{q}{p} } ( \log^{2}{x} + 1 )^{q} } d m ( x ) }_{\text{Lebesgue-Integral}} = \underbrace{ \int\limits_{0}^{\infty} \frac{1}{x^{ \frac{q}{p} } ( \log^{2}{x} + 1 )^{q} } d x }_{\text{Riemann-Integral}} \end{align*}
I was thinking to use the following: Let a decreasing function $\displaystyle{g : ( 0 , \infty ) \longrightarrow \mathbb{R}}$. Then: \begin{align*} \sum\limits_{n=1}^{\infty} 2^{n} g ( 2^{n} ) < \infty \Longleftrightarrow \sum\limits_{n=1}^{\infty} g ( n ) < \infty \Longleftrightarrow \int\limits_{0}^{\infty} g ( x ) d x < \infty. \end{align*}
Since $f$ is decreasing function, it is enough to check the following: \begin{align*} \int\limits_{0}^{\infty} \frac{1}{ ( 2^{x} )^{ \frac{q}{p} } ( \log^{2}{2^x} + 1 )^{q} } d x = \int\limits_{0}^{\infty} \frac{1}{ ( 2^{x} )^{ \frac{q}{p} } ( x^{2} \log^{2}{2} + 1 )^{q} } d x \end{align*}
My issue is that: I cannot show $f \notin L^{q}$. Can you help me? If I have somewhere a mistake please correct me.