0
$\begingroup$

I have the Integral: $$\int^\infty_{20}\frac{1}{x\cdot \ln^{15}(x)}\,dx$$ I know that $$\lim_{x\to \infty}(\ln(x)) = \infty$$

Subsequently, I could substitute with $$\ln(x)$$ in the denominator and try to prove that this integral is convergent because $$f(x) > g(x)$$ But it does not seem to be working. Because it seems divergent and could not state the same for a smaller function. Could you suggest another substitution.

$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Hint: The integral $$ \int_{20}^\infty\frac{dx}{x(\ln(x))^{15}} $$ can be rewritten as $$ \lim_{t\rightarrow\infty}\int_{20}^t\frac{dx}{x(\ln(x))^{15}}. $$ Then, use the substitution $u=\ln(x)$ to get that this integral is $$ \lim_{t\rightarrow\infty}\int_{\ln(20)}^{\ln(t)}\frac{du}{u^{15}}. $$

$\endgroup$
1
  • $\begingroup$ Thank you, it seems easy. The problem was that I did not know that I am able to do the transition between the second and the third step, so I tried another way. $\endgroup$
    – Lazov
    Mar 20, 2016 at 22:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .