Convergence of the integral $\int _{1}^{\infty} \frac{\ln^5(x)}{x^2}dx$

Question

How can I show (with one of the tests for convergence , not by solving) that the integral $$\int _{1}^\infty\frac{\ln^5(x)}{x^2}dx$$ converges?

Answer 1

Abel's theorem: let $g = fh$ be defined on $[a, b)$ with $b \in (a, \infty]$. If it holds that:

• $h$ is continuous, positive, decreasing and $\lim_{x \to b} h(x) = 0$

• $f$ is locally integrable on $[a, b)$ (i.e. integrable in the neighbourhood of any point) and $F$ is bounded on $[a, b)$, where $F: [a,b) \to \Bbb R$,

$$F(x) = \int_a^x f(t) dt$$

Then, $\int_a^b g(x)dx$ is convergent.

Use this with $a = 1$, $b = \infty$, $h(x) = 1/x^2$

Answer 2

Since $\ln^5(x)$ grows more slowly than $x^{1/2}$, you could use a comparison test. For some positive $a>1$, we have:

$$\int_a^\infty \frac{\ln^5(x)}{x^2} dx \leq \int_a^\infty \frac{x^{1/2}}{x^2} dx = \int_a^\infty \frac{dx}{x^{3/2}}$$

You should know that the last expression is convergent by previous results.

Answer 3

If you really meant: $$ \int_1^\infty \frac{\ln(x)^5}{x^2}dx, $$ then observe that for $x$ sufficiently large, $\frac{\ln(x)^5}{\sqrt{x}}$ is less than $1$. In other words, for some $c>1$, $\ln(x)^5<\sqrt{x}$ for all $x>c$. Therefore, the tail of this integral is bounded above by $$ \int_c^\infty\frac{dx}{x^{3/2}}, $$ which converges. Then, $$ \int_1^\infty \frac{\ln(x)^5}{x^2}dx\leq \int_1^c \frac{\ln(x)^5}{x^2}dx+\int_c^\infty\frac{dx}{x^{3/2}}<\infty. $$

Answer 4

Show that $ \lim_{x \to \infty} \frac{\ln^5(x)}{\sqrt{x}}=0$. Hence there is $a>1$ sucht that

$0 < \frac{\ln^5(x)}{\sqrt{x}} \le 1$ for $x>a$. It follows:

$0 < \frac{\ln^5(x)}{x^2} \le \frac{1}{x^{3/2}}$ for $x>a$.

Can you proceed ?

Answer 5

For any $a, b >0$, $(\log x)^a/x^b \to 0$ as $x \to \infty$.

Therefore $\int_1^{\infty}\dfrac{(\log x)^adx}{x^{1+b}} $ converges for $a,b >0$ (and diverges for $b=0$).

Answer 6

We know $$\ln u=\int_1^u\dfrac{1}{t}dt\leq\int_1^u\dfrac{t^\alpha}{t}dt=\dfrac{u^\alpha-1}{\alpha}<\dfrac{u^\alpha}{\alpha}$$ for all $\alpha>0$ so with $\alpha=\dfrac{1}{10}$: $$\int _{1}^\infty\frac{\ln^5(x)}{x^2}dx \leqslant 10^5\int _{1}^\infty\frac{1}{x^\frac32}dx<\infty$$

Answer 7

$\displaystyle \mathcal{I}=\int _{1}^\infty\frac{\ln^5(x)}{x^2}dx$

As $\displaystyle \lim_{x\to +\infty}\frac{\ln^5(x)}{\sqrt{x}}=0\implies \exists a>0, \text{such that } \forall x>a \quad \frac{\ln^5(x)}{\sqrt{x}}<1\iff\frac{\ln^5(x)}{x^2}<\frac{1}{x^{\frac{3}{2}}}$

$\displaystyle\implies \frac{\ln^5(x)}{x^2} =\mathcal{O}\left(\frac{1}{x^{\frac{3}{2}}}\right)\implies \int _{a}^\infty\frac{\ln^5(x)}{x^2}dx$ converges

As $\displaystyle \int _{1}^a\frac{\ln^5(x)}{x^2}\;dx$ is bounded $\implies\mathcal{I}$ converges