# $f \in L_p(1,\infty ))$ if and only if $p \geq 2$

If $$f(x)=\frac{1}{\sqrt{x} (1+\ln{x})}$$ for $$x \in (1, \infty)$$.

I need to show that $$f \in L^p(1, \infty))$$ if and only if $$p \geq 2$$

I suppose that $$p \geq 2$$ and so $$||f||_p^p=\int_1^{\infty}\frac{x^{-p/2}}{(1+\ln{x})^p}dx$$ When I substitute $$1+\ln{x} =u$$ the integral becomes similar to gamma function with negative argument which means its not convergent. I don't know if this is the correct argument or there is another comparison method to prove the convergence of the integral

• check your bounds! remember, $\Gamma(z)$ starts at $0$, your integral will start at $1$ after your substitution Apr 28, 2021 at 20:44
• @NickCastillo you are right! but if the gamma integral exists then so is the integral. Apr 28, 2021 at 20:46
• forgetting about the Gamma function, I got one direction assuming $p\geq2$ Apr 28, 2021 at 20:48
• the only singularity of the integrand is at $u=0$ which is not part of the region of integration Apr 28, 2021 at 20:48

Notice that

$$|f(x)|^p=\frac{1}{x^{p/2}(1+\log(x))^p}\leq\frac{1}{x^{p/2}}$$ for $$x\geq 1$$. This is enough for convergence when $$p>2$$

When $$p=2$$ one gets $$\int^\infty_1\frac{dx}{x(1+\log x)^2}=\int^\infty_1\frac{du}{u^2}=1$$

For $$p<2$$, fix $$\varepsilon>0$$ small enough so that $$\frac{p}{2}+\varepsilon<1$$. Since $$\lim_{x\rightarrow\infty}\frac{1+\log(x)}{x^{\varepsilon/p}}=0$$, one has that $$|f(x)|^p=\frac{1}{x^{p/2}(1+\log(x))^p}\geq\frac{1}{x^{\tfrac{p}{2}+\varepsilon}}$$ for $$x$$ large enough. That will give you divergence for $$p<2$$.

For $$p<2$$, you can compare to an appropriate $$x^q$$ where $$q<-1$$.

For $$p>2$$, you can compare to an appropriate $$x^q$$ where $$q>-1$$.

For $$p=2$$ you can just do the integral of $$f^2$$ by substitution. (This is really the case that is interesting; other than this case, this function is the same as $$\frac{1}{\sqrt{x}}$$ in terms of integrability on $$[1,\infty)$$.