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Test the convergence of the following integral$$\int_{e^2}^\infty {dx\over x\log\log x}$$

I understand that the problem is only at $\infty$ how to proceed ?

Thomas Andrews
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Aman Mittal
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  • Another way is to substitute $u = \log x$, and then you get $$\int_{e^2}^{e^T} \frac{dx}{x\log\log x} = \int_2^T \frac{du}{\log u} = \operatorname{Li}(T).$$ – Daniel Fischer Jul 27 '13 at 17:56

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Hint: Your function $\frac{1}{x\log\log x}$ goes to $0$ more slowly than $\frac{1}{x\log x}$. You can comfortably find $\int_{e^2}^M \frac{dx}{x\log x}$ explicitly, and show that it (sedately) blows up as $M\to \infty$.

André Nicolas
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  • so what you are telling is to use ${1\over {x \log \log x}} < {1 \over {x\log x}}$ ? – Aman Mittal Jul 27 '13 at 17:13
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    The inequality runs the other way, and that's what I am suggesting you use. – André Nicolas Jul 27 '13 at 17:15
  • one side question, for someone who comes across the question for the first time how will he know what function to assume ? like how did you know that we use $1\over x\log x$ ? Is there some use of just practice and memory ? – Aman Mittal Jul 27 '13 at 17:17
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    I this case, I also figured out $\frac{1}{x\log x}$. I was trying to figure out how to do the original integral, and thought, "this would be a lot easier if it was $\frac{1}{x\log x}$, and then realized that was enough :) @AmanMittal – Thomas Andrews Jul 27 '13 at 17:21
  • For an integral of a positive decreasing function, the issue is how fast it goes down. You may already have seen that $\frac{1}{x\log x}$ "doesn't go down fast enough." It helps to have a stock of examples in memory. So it keeps getting easier! There are other ways, like substituting $w$ for $\log\log x$, or for $\log x$, that bring us closer to familiar functions. – André Nicolas Jul 27 '13 at 17:23
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The integral converges iff the series

$$\sum_{n>e^2}\frac1{n\log\log n}\;\;\text{converges}$$

Applying the condensation test to this series, we get it converges iff

$$\sum_{n>e^2}\frac{2^n}{2^n\log\log2^n}=\sum_{n>e^2}\frac1{\log n+\log\log2}$$

and this last series diverges...

DonAntonio
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For $x>100$ $$\frac{dx}{x\ln\ln x} \ge \frac{dx}{x\ln x\ln\ln x}=\frac{d\ln x}{\ln x\ln \ln x}=\frac{d \ln \ln x}{\ln \ln x}=d \ln \ln \ln x,$$which implies the divergence, because the limits of integration are $[\ln \ln \ln 100,\infty)$.

TZakrevskiy
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