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I would like to receive some help with checking the calculation result:

I was calculating the result of the following improper integral:

$$\int\limits_3^\infty \frac{dx}{x \cdot \ln x \cdot (\ln\ln x)^{1 + \alpha}} \, \,\,\text{where}\,\,\, \alpha > 0$$

I did it this way:

$$\int\limits_3^\infty \frac{dx}{x \cdot \ln x \cdot (\ln\ln x)^{1 + \alpha}} = \lim_{\beta \to \infty} \int\limits_3^{\beta} \frac{dx}{x \cdot \ln x \cdot (\ln\ln x)^{1 + \alpha}} = (*).$$

As i understand, now we should calculate the definite integral $\int\limits_3^{\beta} \frac{dx}{x \cdot \ln x \cdot (\ln \ln x)^{1 + \alpha}}$,using $\beta$ as constant. Now i used the next substitution: $u(x) = \ln(\ln x)$, $du(x) = \frac{dx}{x \ln x}$. I calculated the new boundaries: $u(3) = \ln(\ln3)$, $u(\beta) = \ln(\ln \beta)$.

$$(*) = \lim_{\beta \to \infty} \int\limits_3^{\beta} \ln(\ln x)^{-1-\alpha} \frac{dx}{x \ln x} = \lim_{\beta \to \infty} \int\limits_{\ln(\ln3)}^{\ln(\ln \beta)} u^{-1 - \alpha}du.$$

From here i got the next result :

$$ \lim_{\beta \to \infty} \left( \frac{1}{\alpha} \left( \frac{1}{\ln(\ln(\ln(\ln3)))^{\alpha}} - \frac{1}{\ln(\ln(\ln(\ln \beta)))^{\alpha}}\right) \right) = \frac{1}{\alpha} \cdot \frac{1}{\ln(\ln(\ln(\ln 3)))^{\alpha}}.$$

Please, could you tell me if my calculation was right?

Thank you, for your time and help!

John Glenn
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MathsLearner
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  • Too many logarithms. At the end, you seem to have used the substitution a second time for the integral limits. $$\int_{\xi}^{\eta} u^{-1-\alpha},du = \frac{1}{\alpha}\biggl(\frac{1}{\xi^{\alpha}} - \frac{1}{\eta^{\alpha}}\biggr)$$ for $0 < \xi, \eta$. – Daniel Fischer Apr 04 '18 at 14:49
  • @ Daniel Fischer : Maybe the more steps will be more helpful: $\lim_{\beta \to \infty} \int\limits_{\ln (\ln 3)}^{\ln (\ln \beta)} u^{-1 - \alpha} du = \lim_{\beta \to \infty} \frac{u^{-\alpha}}{-\alpha}|{\ln(\ln 3)}^{\ln(\ln \beta)} = \lim{\beta \to \infty} \frac{(\ln \ln x)^{-\alpha}}{-\alpha}|{\ln (\ln 3)}^{\ln(\ln \beta)} = \lim{\beta \to \infty} (\frac{1}{(-\alpha)(\ln \ln \ln \ln \beta)^{\alpha}} - \frac{1}{(-\alpha)(\ln \ln \ln \ln 3)^{\alpha}})$ – MathsLearner Apr 04 '18 at 15:26
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    Ah. When you substitute back and replace $u$ with $\ln (\ln x)$, then you also must substitute back the integral limits. You get $$\lim_{\beta \to \infty} \frac{(\ln \ln x)^{-\alpha}}{-\alpha}\biggr\rvert_{3}^{\beta},.$$ – Daniel Fischer Apr 04 '18 at 15:30
  • @ Daniel Fischer: Yes. Now i see where i made a mistake! I thought that i knew what to do with new boundaires, but i forgot few stuff. :) Thank you very much! – MathsLearner Apr 04 '18 at 15:39

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