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How can I prove that $$\int_0^\infty \frac{1}{x^\alpha(1+\log^2x)^\alpha}dx$$ diverges for any $\alpha>0$?

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    i would take a look at a right neigbourhood of $x=0$ and a neighbourhood of $x=\infty$ – 2017-02-01

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Actually the integral converges if $\alpha=1$, but only then.

To prove this, decompose the integral dyadically to see

$$\int_0^\infty \frac1{x^\alpha (1+\log^2 x)^\alpha} dx = \sum_{k\in\mathbb{Z}} \int_{2^k}^{2^{k+1}} \frac1{x^\alpha (1+\log^2 x)^\alpha} dx $$

Note that $\int_{2^k}^{2^{k+1}} \frac1{x^\alpha (1+\log^2 x)^\alpha} dx $ is bounded from above and below (up to constants independent of $k$) by $$\frac{2^k }{2^{k\alpha} (1+k^2)^\alpha}$$

Thus, the integral converges if and only if the sum $$\sum_{k\in\mathbb{Z}} \frac{2^{k(1-\alpha)}}{(1+k^2)^{\alpha}}$$

converges.

Clearly, this is the case if and only if $\alpha=1$, since if $\alpha < 1$ the portion where $k$ tends to $+\infty$ diverges and if $\alpha > 1$ the portion where $k$ tends to $-\infty$ diverges.

Remark. This procedure is referred to as Cauchy condensation test.

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    Thanks for your answer. Could you add a precise statement of the Cauchy condensation test that you are applying and a remark on where you're applying it? Also, it is possible to prove our result using only the comparison test? If yes, how? – 2017-02-01
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    Please read what I wrote. I gave you a complete proof of the test I'm applying (and it uses only the comparison test). Just wanted to let you know that this is called "Cauchy condensation test". – 2017-02-01
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    I moved the reference to the Cauchy condensation test that confused you to the end. It is really unrelated to the logic of the argument. – 2017-02-01
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    Thanks, it clears that up. Could you add some details on that bound "up to a constant"? – 2017-02-01
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    If $2^k\le x\le 2^{k+1}$, then $c 2^{k\alpha} (1+k^2)^\alpha\le x^\alpha (1+\log^2 x)^\alpha \le C 2^{k\alpha} (1+k^2)^\alpha$, where $c,C$ are some constants (you can try to see what they are, but it's not important) not depending on $k$. – 2017-02-01
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    And how do you get $$\frac{2^k }{2^{k\alpha} (1+k^2)^\alpha}?$$ – 2017-02-01
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    The denominator should be clear, it comes from estimating $x^\alpha (1+\log^2 x)^\alpha$ as in my last comment. The numerator comes from the length of the interval we are integrating over: $x\in [2^k, 2^{k+1}]$. – 2017-02-02