Lebesgue integration criteria - Verify integrability

Question

I have the following definite integral: $$\int_{2}^{\infty}\frac{1}{x^a(\ln x)^2}\,dx$$ I want to discuss the integrability of this function at $+\infty $ for the different values of the real parameter $a$. I know that $$\frac{1}{x^a(\ln x)^2}\in L_1(]2, \infty[)$$ for $a\ge 1$. I have tried using the asymptotic expressions and the fact that $(\ln x)^2 = o(x^2)$, but it didn't help. Finally I tried to prove it using limits, but again I didn't manage. I guess I am missing something. Any ideas?

Answer 1

Hint. You may use the comparison test for improper integrals.

For $a\geq 1$ and $x\geq 2$, $$0<\frac{1}{x^a(\ln x)^2}\leq \frac{1}{x(\ln x)^2}$$ and $$\int_2^{\infty} \frac{1}{x(\ln x)^2},dx=\left[-\frac{1}{\ln x}\right]_2^{+\infty}=\frac{1}{\ln 2}.$$ On the other hand, for $a<1$, $$\lim_{x\to +\infty}\frac{x^{(a+1)/2}}{x^a(\ln x)^2}=\lim_{x\to +\infty}\frac{x^{(1-a)/2}}{(\ln x)^2}=\lim_{t\to +\infty}\frac{e^{(1-a)t/2}}{t^2}=+\infty.$$ Hence, for $x$ sufficiently large, $$\frac{1}{x^a(\ln x)^2}\geq \frac{1}{x^{(a+1)/2}}$$ and, $$\int_2^{\infty} \frac{dx}{x^{(a+1)/2}}=\left[\frac{x^{(1-a)/2}}{(1-a)/2}\right]_2^{+\infty}=+\infty.$$

Answer 2

The substitution $y=\ln x$ converts the integral to $\int_{\ln 2}^{\infty}\frac {e^{-(a-1)x}} {x^{2}} dx$. can you see from here that the integral converges if and only if $a \geq 1$?

Answer 3

Actually, using the comparison theorem, you can prove that$$\int_2^{+\infty}\frac{\mathrm d x}{x^a\,(\ln x)^b}\quad\text{converges if} \quad\begin{cases}a>1,\;\text{or}\\ a=1\;\text{ and }\;b>1,\end{cases}$$ and diverges in all other cases.

Some details: suppose $b>0$.

We have $\;0<\dfrac1{x^a\,(\ln x)^b}<\dfrac1{x^a}\;$ for all $x>\mathrm e$, so it converges if $a>1$.

If $a=1$ and $b>1$, we have $$\int_2^{+\infty}\frac{\mathrm d x}{x\,(\ln x)^b}=-\frac1{(b-1)(\ln x)^{b-1}}\Biggr\vert_2^\infty=\frac1{(b-1)(\ln 2)^{b-1}}$$ If $a<1$, we have $\;(\ln x)^b=o(x^h)$ for any $h>0$, so $$\frac1{x^a\,(\ln x)^b}>\frac1{x^a\, x^h}=\frac1{x^{a+h}}\quad\text{if $\,x\,$ is large enough},$$ and you prove it diverges choosing $h$ such that $a+h<1$.