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

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

2 Answers 2


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 $[1,\infty)$.


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